Financial Mathematics (2003) Richard F Bass

Financial Mathematics

Spring 2003

Richard F. Bass

Department of Mathematics

University of Connecticut

These notes are c

2003 by Richard Bass. They may be used for personal use or

class use, but not for commercial purposes.

0. Introduction.

In this course we will study mathematical finance. Mathematical finance is not

about predicting the price of a stock. What it is about is figuring out the price of options

and derivatives.

The most familiar type of option is the option to buy a stock at a given price at

a given time. For example, suppose Microsoft is currently selling today at $40 per share.

A European call option is something I can buy that gives me the right to buy a share of

Microsoft at some future date. To make up an example, suppose I have an option that

allows me to buy a share of Microsoft for $50 in three months time, but does not compel

me to do so. If Microsoft happens to be selling at $45 in three months time, the option is

worthless. I would be silly to buy a share for $50 when I could call my broker and buy it

for $45. So I would choose not to exercise the option. On the other hand, if Microsoft is

selling for $60 three months from now, the option would be quite valuable. I could exercise

the option and buy a share for $50. I could then turn around and sell the share on the

open market for $60 and make a profit of $10 per share. Therefore this stock option I

possess has some value. There is some chance it is worthless and some chance that it will

lead me to a profit. The basic question is: how much is the option worth today?

The huge impetus in financial derivatives was the seminal paper of Black and Scholes

in 1973. Although many researchers had studied this question, Black and Scholes gave a

definitive answer, and a great deal of research has been done since. These are not just

academic questions; today the market in financial derivatives is larger than the market

in stock securities. In other words, more money is invested in options on stocks than in

stocks themselves.

Options have been around for a long time. The earliest ones were used by manu-

facturers and food producers to hedge their risk. A farmer might agree to sell a bushel of

wheat at a fixed price six months from now rather than take a chance on the vagaries of

market prices. Similarly a steel refinery might want to lock in the price of iron ore at a

fixed price.

The sections of these notes can be grouped into 5 categories. The first is elementary

probability. Although someone who has had a course in undergraduate probability will

be familiar with some of this, we will talk about a number of topics that are not usually

covered in such a course: σ-fields, conditional expectations, martingales.

The second

category is the binomial asset pricing model. This is just about the simplest model of

a stock that one can imagine, and this will provide a case where we can see most of

the major ideas of mathematical finance, but in a very simple setting.

Then we will

turn to advanced probability, that is, ideas such as Brownian motion, stochastic integrals,

stochastic differential equations, Girsanov tranformation. Although to do this rigorously

requires measure theory, we can still learn enough to understand and work with these

concepts. We then return to finance and work with the continuous model. We will derive

the Black-Scholes formula, see the Fundamental Theorem of Asset Pricing, work with

equivalent martingale measures, and the like. The fifth main category is term structure

models, which means models of interest rate behavior.

1. Review of elementary probability.

Let’s begin by recalling some of the definitions and basic concepts of elementary

probability. We will only work with discrete models at first.

We start with an arbitrary set, called the probability space, which we will denote

by Ω, the Greek letter “capital omega.” We are given a class F of subsets of Ω. These are

called events. We require F to be a σ-field. This means that

(1) ∅ ∈ F ,

(2) Ω ∈ F ,

(3) A ∈ F implies A

∈ F , and

(4) A

, A

, . . . ∈ F implies both ∪

∞
i=1

∈ F and ∩

∞
i=1

∈ F .

Here A

= {ω ∈ Ω : ω /

∈ A} denotes the complement of A. ∅ denotes the empty set, that

is, the set with no elements. We will use without special comment the usual notations of

∪ (union), ∩ (intersection), ⊂ (contained in), ∈ (is an element of).

Typically, in an elementary probability course, F will consist of all subsets of

Ω, but we will later need to distinguish between various σ-fields.

Here is an exam-

ple. Suppose one tosses a coin two times and lets Ω denote all possible outcomes. So

Ω = {HH, HT, T H, T T }. A typical σ-field F would be the one consisting of all subsets (of

which there are 16). In this case it is trivial to show that F is a σ-field, since every subset

is in F . But if we let G = {∅, Ω, {HH, HT }, {T H, T T }}, then G is also a σ-field. One has

to check the definition, but to illustrate, the event {HH, HT } is in G, so we require the

complement of that set to be in G as well. But the complement is {T H, T T } and that

event is indeed in G.

One point of view which we will explore much more fully later on is that the σ-

field tells you what events you know. In this example, F is the σ-field where you “know”

everything, while G is the σ-field where you “know” only the result of the first toss but not

the second. We won’t try to be precise here, but to try to add to the intuition, suppose

one knows whether an event in F has happened or not for a particular outcome. We

would then know which of the events {HH}, {HT }, {T H}, or {T T } has happened and so

would know what the two tosses of the coin showed. On the other hand, if we know which

events in G happened, we would only know whether the event {HH, HT } happened, which

means we would know that the first toss was a heads, or we would know whether the event

{T H, T T } happened, in which case we would know that the first toss was a tails. But

there is no way to tell what happened on the second toss from knowing which events in G

happened. Much more on this later.

The third basic ingredient is a function P on F satisfying

(1) if A ∈ F , then 0 ≤ P(A) ≤ 1,
(2) P(Ω) = 1, and
(3) if A

, A

, . . . ∈ F are pairwise disjoint, then P(∪

∞
i=1

) =

∞
i=1

P(A

P is called a probability or probability measure.

There are a number of conclusions one can draw from this definition.

As one

example, if A ⊂ B, then P(A) ≤ P(B) and P(A

) = 1 − P(A).

Someone who has had real analysis will realize that a σ-field is the same thing as a

σ-algebra and a probability is a measure of total mass one.

A random variable (abbreviated r.v.) is a function X from Ω to R, the reals. To

be more precise, to be a r.v. X must also be measurable, which means that {ω : X(ω) ≥

a} ∈ F for all reals a.

The notion of measurability has a simple definition but is a bit subtle. If we take

the point of view that we know all the events in G, then if Y is G-measurable, then we

know Y .

Here is an example. In the example where we tossed a coin two times, let X be the

number of heads in the two tosses. Then X is F measurable but not G measurable. To

see this, let us consider A

= {ω ∈ Ω : X(ω) ≥ a}. This event will equal











Ω

if a ≤ 0;

{HH, HT, T H}

if 0 < a ≤ 1;

{HH}

if 1 < a ≤ 2;

∅

if 2 < a.

For example, if a =

3
2

, then the event where the number of heads is

3
2

or greater is the

event where we had two heads, namely, {HH}. Now observe that for each a the event A

is in F because F contains all subsets of Ω. Therefore X is measurable with respect to

F . However it is not true that A

is in G for every value of a – take a =

3
2

as just one

example. So X is not measurable with respect to the σ-field G.

A discrete r.v. is one where P(ω : X(ω) = a) = 0 for all but countably many a’s.

In defining sets one usually omits the ω; thus (X = x) is the same as {ω : X(ω) = x}.

In the discrete case, to check measurability with respect to a σ-field F , it is enough

that (X = a) ∈ F for all reals a. The reason for this is if x

, x

, . . . are the values of x

for which P(X = x) 6= 0, then we can write (X ≥ a) = ∪

≥a

(X = x

) and we have a

countable union. So if (X = x

) ∈ F , then (X ≥ a) ∈ F .

Given a discrete r.v. X, the expectation or mean is defined by

E X =

xP(X = x)

provided the sum converges. If X only takes finitely many values, then this is a finite sum

and of course it will converge. This is the situation that we will consider for quite some

time. However, if X can take an infinite number of values (but countable), convergence

needs to be checked.

There is an alternate definition of expectation which is equivalent in the discrete

setting. Set

E X =

ω∈Ω

X(ω)P({ω}).

To see that this is the same, we have

xP(X = x) =

{ω∈Ω:X(ω)=x}

P({ω})

{ω∈Ω:X(ω)=x}

X(ω)P({ω})

ω∈Ω

X(ω)P({ω}).

The advantage of the second definition is that some properties of expectation, such as

E (X + Y ) = E X + E Y , are immediate, while with the first definition they require quite
a bit of proof.

Two events A and B are independent if P(A ∩ B) = P(A)P(B). Two random

variables X and Y are independent if P(X ∈ A, Y ∈ B) = P(X ∈ A)P(X ∈ B) for all
A and B that are subsets of the reals. The comma in the expression on the left hand

side means “and.” The extension of this definition to the case of more than two events or

random variables is obvious.

Two σ-fields F and G are independent if A and B are independent whenever A ∈ F

and B ∈ G. A r.v. X and a σ-field G are independent if P((X ∈ A) ∩ B) = P(X ∈ A)P(B)
whenever A is a subset of the reals and B ∈ G.

As an example, suppose we toss a coin two times and we define the σ-fields G

{∅, Ω, {HH, HT }, {T H, T T }} and G

= {∅, Ω, {HH, T H}, {HT, T T }}. Then G

and G

are

independent if P(HH) = P(HT ) = P(T H) = P(T T ) =

1
4

. (Here we are writing P(HH)

when a more accurate way would be to write P({HH}).) An easy way to understand this
is that if we look at an event in G

that is not ∅ or Ω, then that is the event that the first

toss is a heads or it is the event that the first toss is a tails. Similarly, a set other than ∅

or Ω in G

will be the event that the second toss is a heads or that the second toss is a

tails.

If two r.v.s X and Y are independent, we have the multiplication theorem, which

says that E (XY ) = (E X)(E Y ) provided all the expectations are finite.

Suppose X

, . . . , X

are n independent r.v.s, such that for each one P(X

= 1) = p,

P(X

= 0) = 1 − p, where p ∈ [0, 1]. The random variable S

n
i=1

is called a

binomial r.v., and represents, for example, the number of successes in n trials, where the

probability of a success is p. An important result in probability is that

P(S

= k) =

k!(n − k)!

(1 − p)

n−k

We close this section with a definition of conditional probability. The probability

of A given B, written P(A | B) is defined by

P(A ∩ B)

P(B)

provided P(B) 6= 0. The conditional expectation of X given B is defined to be

E [X; B]

P(B)

provided P(B) 6= 0. The notation E [X; B] means E [X1

], where 1

(ω) is 1 if ω ∈ B and

0 otherwise. Another way of writing E [X; B] is

E [X; B] =

ω∈B

X(ω)P({ω}).

2. Conditional expectation.

Suppose we have 200 men and 100 women, 70 of the men are smokers, and 50 of

the women are smokers. If a person is chosen at random, then the conditional probability

that the person is a smoker given that it is a man is 70 divided by 200, or 35%, while the

conditional probability the person is a smoker given that it is a women is 50 divided by

100, or 50%. We will want to be able to encompass both facts in a single entity.

The way to do that is to make conditional probability a random variable rather

than a number. To reiterate, we will make conditional probabilities random. Let M, W be

man, woman, respectively, and S, S

smoker and nonsmoker, respectively. We have

P(S | M ) = .35,

P(S | W ) = .50.

We introduce the random variable

(.35)1

+ (.50)1

and use that for our conditional probability. So on the set M its value is .35 and on the

set W its value is .50.

We need to give this random variable a name, so what we do is let G be the σ-field

consisting of {∅, Ω, M, W } and denote this random variable P(S | G). Thus we are going
to talk about the conditional probability of an event given a σ-field.

What is the precise definition? Suppose there exist finitely (or countably) many sets

, B

, . . ., all having positive probability, such that they are pairwise disjoint, Ω is equal

to their union, and G is the σ-field one obtains by taking all finite or countable unions of

the B

. Then the conditional probability of A given G is

P(A | G) =

P(A ∩ B

)

P(B

)

(ω).

In short, on the set B

the conditional probability is equal to P(A | B

Not every σ-field can be so represented, so this definition will need to be extended

when we get to continuous models. σ-fields that can be represented in this way are called

finitely (or countably) generated and are said to be generated by the sets B

, B

, . . ..

Let’s look at another example. Suppose Ω consists of the possible results when we

toss a coin three times: HHH, HHT, etc. Let F

denote all subsets of Ω. Let F

consist of

the sets ∅, Ω, {HHH, HHT, HT H, HT T }, and {T HH, T HT, T T H, T T T }. So F

consists

of those events that can be determined by knowing the result of the first toss. We want to

let F

denote those events that can be determined by knowing the first two tosses. This will

include the sets ∅, Ω, {HHH, HHT }, {HT H, HT T }, {T HH, T HT }, {T T H, T T T }. This is

not enough to make F

a σ-field, so we add to F

all sets that can be obtained by taking

unions of these sets.

Suppose we tossed the coin independently and suppose that it was fair. Let us

calculate P(A | F

), P(A | F

), and P(A | F

) when A is the event {HHH}.

First

the conditional probability given F

. Let C

= {HHH, HHT, HT H, HT T } and C

{T HH, T HT, T T H, T T T }. On the set C

the conditional probability is P(A∩C

)/P(C

) =

P(HH H)/P(C

) =

1
8

1
2

1
4

. On the set C

the conditional probability is P(A∩C

)/P(C

)

= P(∅)/P(C

) = 0. Therefore P(A | F

) = (.25)1

. This is plausible – the probability of

getting three heads given the first toss is

1
4

if the first toss is a heads and 0 otherwise.

Next let us calculate P(A | F

). Let D

= {HHH, HHT }, D

= {HT H, HT T }, D

= {T HH, T HT }, D

= {T T H, T T T }. So F

is the σ-field consisting of all possible unions

of some of the D

’s. P(A | D

) = P(HHH)/P(D

) =

1
8

1
4

1
2

. Also, as above, P(A |

) = 0 for i = 2, 3, 4. So P(A | F

) = (.50)1

. This is again plausible – the probability

of getting three heads given the first two tosses is

1
2

if the first two tosses were heads and

0 otherwise.

What about conditional expectation? Given a random variable X, we define

E [X | G] =

E [X; B

]

P(B

)

This is the obvious definition, and it agrees with what we had before because E [1

| G] =

P(A | G).

We now turn to some properties of conditional expectation. Some of the following

propositions may seem a bit technical. In fact, they are! However, these properties are

crucial to what follows and there is no choice but to master them.

Proposition 2.1. E [X | G] is G measurable, that is, if Y = E [X | G], then (Y > a) is a
set in G for each real a.

Proof. By the definition,

Y = E [X | G] =

E [X; B

]

P(B

)

if we set b

= E [X; B

]/P(B

). The set (Y ≥ a) is a union of some of the B

, namely, those

for which b

≥ a. But the union of any collection of the B

is in G.

An example might help. Suppose

Y = 2 · 1

+ 3 · 1

+ 6 · 1

+ 4 · 1

and a = 3.5. Then (Y ≥ a) = B

∪ B

, which is in G.

Proposition 2.2. If C ∈ G and Y = E [X | G], then E [Y ; C] = E [X; C].

Proof. Since Y =

E [X;B

]

P(B

)

and the B

are disjoint, then

E [Y ; B

] =

E [X; B

]

P(B

)

E 1

= E [X; B

Now if C = B

∪ · · · ∪ B

∪ · · ·, summing the above over the j

gives E [Y ; C] = E [X; C].

Let us look at the above example for this proposition, and let us do the case where

C = B

. Note 1

= 1

because the product is 1 · 1 = 1 if ω is in B

and 0 otherwise.

On the other hand, it is not possible for an ω to be in more than one of the B

, so

= 0 if i 6= 2. Mutiplying Y in the above example by 1

, we see that

E [Y ; C] = E [Y ; B

] = E [Y 1

] = E [3 · 1

]

= 3E [1

] = 3P(B

However the number 3 is not just any number; it is E [X; B

]/P(B

). So

3P(B

) =

E [X; B

]

P(B

)

P(B

) = E [X; B

] = E [X; C],

just as we wanted. If C = B

∪ B

, for example, we then write

E [X; C] = E [X1

] = E [X(1

+ 1

)]

= E [X1

] + E [X1

] = E [X; B

] + E [X; B

By the first part, this equals E [Y ; B

]+E [Y ; B

], and we undo the above string of equalities

but with Y instead of X to see that this is E [Y ; C].

If a r.v. Y is G measurable, then for any a we have (Y = a) ∈ G which means that

(Y = a) is the union of one of more of the B

. Since the B

are disjoint, it follows that Y

must be constant on each B

Again let us look at an example. Suppose Z takes only the values 1, 3, 4, 7. Let

= (Z = 1), D

= (Z = 3), D

= (Z = 4), D

= (Z = 7). Note that we can write

Z = 1 · 1

+ 3 · 1

+ 4 · 1

+ 7 · 1

To see this, if ω ∈ D

, for example, the right hand side will be 0 + 3 · 1 + 0 + 0, which agrees

with Z(ω). Now if Z is G measurable, then (Z ≥ a) ∈ G for each a. Take a = 7, and we

see D

∈ G. Take a = 4 and we see D

∪ D

∈ G. Taking a = 3 shows D

∪ D

∈ G.

Now D

= (D

∪ D

) ∩ D

, so since G is a σ-field, D

∈ G. Similarly D

, D

∈ G. Because

sets in G are unions of the B

’s, we must have Z constant on the B

’s. For example, if it

so happened that D

= B

, D

= B

∪ B

, D

= B

∪ B

, and D

= B

, then

Z = 1 · 1

+ 3 · 1

+ 4 · 1

+ 3 · 1

+ 7 · 1

+ +4 · 1

+ 4 · 1

We still restrict ourselves to the discrete case. In this context, the properties given

in Propositions 2.1 and 2.2 uniquely determine E [X | G].

Proposition 2.3. Suppose Z is G measurable and E [Z; C] = E [X; C] whenever C ∈ G.
Then Z = E [X | G].

Proof. Since Z is G measurable, then Z must be constant on each B

. Let the value of Z

on B

be z

. So Z =

. Then

P(B

) = E [Z; B

] = E [X; B

or z

= E [X; B

]/P(B

) as required.

The following propositions contain the main facts about this new definition of con-

ditional expectation that we will need.

Proposition 2.4. (1) If X

≥ X

, then E [X

| G] ≥ E [X

| G].

(2) E [aX

+ bX

| G] = aE [X

| G] + bE [X

| G].

(3) If X is G measurable, then E [X | G] = X.
(4) E [E [X | G]] = E X.
(5) If X is independent of G, then E [X | G] = E X.

If we were to think of E [X | G] as the best prediction of X given G, we can give an

interpretation of (1)-(5). (1) says that if X

is larger than X

, then the predicted value of

should be larger than the predicted value of X

. (2) says that the predicted value of

+ X

should be the sum of the predicted values. (3) says that if we know G and X is

G measurable, then we know X and our best prediction of X is X itself. (4) says that the

average of the predicted value of X should be the predicted value of X. (5) says that if

knowing G gives us no additional information on X, then the best prediction for the value

of X is just E X.

Proof. (1) and (2) are immediate from the definition. To prove (3), note that if Z =

X, then Z is G measurable and E [X; C] = E [Z; C] for any C ∈ G; this is trivial. By
Proposition 2.3 it follows that Z = E [X | G]; this proves (3). To prove (4), if we let C = Ω
and Y = E [X | G], then E Y = E [Y ; C] = E [X; C] = E X.

Last is (5). Let Z = E X. Z is constant, so clearly G measurable. By the in-

dependence, if C ∈ G, then E [X; C] = E [X1

] = (E X)(E 1

) = (E X)(P(C)). But

E [Z; C] = (E X)(P(C)) since Z is constant. By Proposition 2.3 we see Z = E [X | G].

Propostion 2.5. If Z is G measurable, then E [XZ | G] = ZE [X | G].

Proof. Note that ZE [X | G] is G measurable, so by Proposition 2.3 we need to show its
expectation over sets C in G is the same as that of XZ. As in the proof of Proposition

2.2, it suffices to consider only the case when C is one of the B

. Now Z is G measurable,

hence it is constant on B

; let its value be z

. Then

E [ZE [X | G]; B

] = E [z

E [X | G]; B

] = z

E [E [X | G]; B

] = z

E [X; B

] = E [XZ; B

]

as desired.

This proposition says that as far as conditional expectations with respect to a σ-

field G go, G-measurable random variables act like constants: they can be taken inside or

outside the conditional expectation at will.

Proposition 2.6. If H ⊂ G ⊂ F , then

E [E [X | H] | G] = E [X | H] = E [E [X | G] | H].

Proof. E [X | H] is H measurable, hence G measurable, since H ⊂ G. The left hand
equality now follows by Proposition 2.4(3). To get the right hand equality, let W be the

right hand expression. It is H measurable, and if C ∈ H ⊂ G, then

E [W ; C] = E [E [X | G]; C] = E [X; C]

as required.

In words, if we are predicting a prediction of X given limited information, this is

the same as a single prediction given the least amount of information.

If Y is a discrete random variables, that is, it takes only countably many values

, y

, . . ., we let B

= (Y = y

). These will be disjoint sets whose union is Ω. If σ(Y )

is the collection of all unions of the B

, then σ(Y ) is a σ-field, and is called the σ-field

generated by Y . It is easy to see that this is the smallest σ-field with respect to which Y

is measurable. We write E [X | Y ] for E [X | σ(Y )].

3. Martingales.

Suppose we have a sequence of σ-fields F

⊂ F

· · ·. An example would be

repeatedly tossing a coin and letting F

be the sets that can be determined by the first

k tosses. Another example is to let F

be the events that are determined by the values

of a stock at times 1 through k. A third example is to let X

, X

, . . . be a sequence of

random variables and let F

be the σ-field generated by X

, . . . , X

, the smallest σ-field

with respect to which X

, . . . , X

are measurable.

A r.v. X is integrable if E |X| < ∞. Given an increasing sequence of σ-fields F

, a

sequence of r.v.’s X

is adapted if X

is F

measurable for each n.

A martingale M

is a sequence of random variables such that M

is integrable for

all n, M

is adapted to F

, and

E [M

n+1

| F

] = M

(3.1)

for each n. Martingales will be ubiquitous in financial math.

The word “martingale” comes from the piece of a horse’s bridle that runs from the

horse’s head to its chest. It keeps the horse from raising its head too high. It turns out

that martingales in probability cannot get too large.

Here is an example. Let X

, X

, . . . be a sequence of independent r.v.’s with mean 0

that are independent. Set F

= σ(X

, . . . , X

), the σ-field generated by X

, . . . , X

. Let

n
i=1

. Then

E [M

n+1

| F

] = X

+ · · · + X

+ E [X

n+1

| F

] = M

+ E X

n+1

= M

where we used the independence.

Another example: suppose in the above that the X

all have variance 1, and let

= S

− n, where S

n
i=1

. We compute

E [M

n+1

| F

] = E [S

+ 2X

n+1

+ X

n+1

| F

] − (n + 1).

We have E [S

| F

] = S

since S

is F

measurable. E [2X

n+1

| F

] = 2S

E [X

n+1

] = 2S

E X

n+1

= 0. And E [X

n+1

| F

] = E X

n+1

= 1. Substituting, we obtain

E [M

n+1

| F

] = M

, or M

is a martingale.

A third example: Suppose you start with a dollar and you are tossing a fair coin

independently. If it turns up heads you double your fortune, tails you go broke. This is

“double or nothing.” Let M

be your fortune at time n. To formalize this, let X

, X

, . . .

be independent r.v.’s that are equal to 2 with probability

1
2

and 0 with probability

1
2

. Then

= X

· · · X

. To compute the conditional expectation, note E X

n+1

= 1. Then

E [M

n+1

| F

] = M

E [X

n+1

| F

] = M

E X

n+1

= M

using the independence.

A final example for now: let F

, F

, . . . be given and let X be a fixed r.v. Let

= E [X | F

]. We have

E [M

n+1

| F

] = E [E [X | F

n+1

] | F

] = E [X | F

] = M

4. Properties of martingales.

When it comes to discussing American options, we will need the concept of stopping

times. A mapping τ from Ω into the nonnegative integers is a stopping time if (τ = k) ∈ F

for each k.

An example is τ = min{k : S

≥ A}. This is a stopping time because (τ = k) =

, . . . , S

k−1

< A, S

≥ A) ∈ F

. We can think of a stopping time as the first time

something happens. σ = max{k : S

≥ A}, the last time, is not a stopping time.

Here is an intuitive description of a stopping time. If I tell you to drive to the city

limits and then drive until you come to the second stop light after that, you know when

you get there that you have arrived; you don’t need to have been there before or to look

ahead. But if I tell you to drive until you come to the second stop light before the city

limits, either you must have been there before or else you have to go past where you are

supposed to stop, continue on to the city limits, and then turn around and come back two

stop lights. You don’t know when you first get to the second stop light before the city

limits that you get to stop there. The first set of instructions forms a stopping time, the

second set does not.

If M

is an adapted sequence of integrable r.v.’s with

E [M

n+1

| F

] ≥ M

for each n, then M

is a submartingale. (Is the “≥” is replaced by “=”, of course, that is

what we called a martingale; if the “≥” is replaced by “≤”, we call M

a supermartingale.)

Our first result is Jensen’s inequality.

Proposition 4.1. If g is convex, then

g(E [X | G]) ≤ E [g(X) | G]

provided all the expectations exist.

For ordinary expectations rather than conditional expectations, this is still true.

That is, if g is convex and the expectations exist, then

g(E X) ≤ E [g(X)].

We already know some special cases of this: when g(x) = |x|, this says |E X| ≤ E |X|;
when g(x) = x

, this says (E X)

≤ E X

, which we know because E X

− (E X)

E (X − E X)

≥ 0.

Proof. If g is convex, then the graph of g lies above all the tangent lines. Even if g does

not have a derivative at x

, there is a line passing through x

which lies beneath the graph

of g. So for each x

there exists c(x

) such that

g(x) ≥ g(x

) + c(x

)(x − x

Apply this with x = X(ω) and x

= E [X | G](ω). We then have

g(X) ≥ g(E [X | G]) + c(E [X | G])(X − E [X | G]).

One can check that c can be chosen so that c(E [X | G]) is G measurable.

Now take the conditional expectation with respect to G. The first term on the right

is G measurable, so remains the same. The second term on the right is equal to

c(E [X | G])E [X − E [X | G] | G] = 0.

One reason we want Jensen’s inequality is to show that a convex function applied

to a martingale yields a submartingale.

Proposition 4.2. If M

is a martingale and g is convex, then g(M

) is a submartingale,

provided all the expectations exist.

Proof. By Jensen’s inequality,

E [g(M

n+1

) | F

] ≥ g(E [M

n+1

| F

]) = g(M

If M

is a martingale, then E M

= E [E [M

n+1

| F

]] = E M

n+1

. So E M

E M

= · · · = E M

. Doob’s optional stopping theorem says the same thing holds when

fixed times n are replaced by stopping times.

Theorem 4.3. Suppose K is a positive integer, N is a stopping time such that N ≤ K

a.s., and M

is a martingale. Then

E M

= E M

Here, to evaluate M

, one first finds N (ω) and then evaluates M

(ω) for that value of N .

Proof. We have

E M

k=0

E [M

; N = k].

If we show that the k-th summand is E [M

; N = k], then the sum will be

k=0

E [M

; N = k] = E M

as desired. We have

E [M

; N = k] = E [M

; N = k]

by the definition of M

. Now (N = k) is F

measurable, so by Proposition 2.2 and the

fact that M

= E [M

k+1

| F

E [M

; N = k] = E [M

k+1

; N = k].

We have (N = k) ∈ F

⊂ F

k+1

. Since M

k+1

= E [M

k+2

| F

k+1

], Proposition 2.2 tells us

that

E [M

k+1

; N = k] = E [M

k+2

; N = k].

We continue, using (N = k) ∈ F

⊂ F

k+1

⊂ F

k+2

, and we obtain

E [M

; N = k] = E [M

k+1

; N = k] = · · · = E [M

; N = k].

If we change the equalities in the above to inequalities, the same result holds for sub-

martingales.

As a corollary we have two of Doob’s inequalities:

Theorem 4.4. (a) If M

is a nonnegative submartingale,

P(max

k≤n

≥ λ) ≤

E M

(b)

E (max

k≤n

) ≤ 4E M

Proof. Set M

n+1

= M

. It is easy to see that the sequence M

, M

, . . . , M

n+1

is also a

martingale. Let N = min{k : M

≥ λ} ∧ (n + 1), the first time that M

is greater than or

equal to λ, where a ∧ b = min(a, b). Then

P(max

k≤n

≥ λ) = P(N ≤ n)

and if N ≤ n, then M

≥ λ. Now

P(max

k≤n

≥ λ) = E [1

(N ≤n)

] ≤ E

; N ≤ n

(4.1)

E [M

N ∧n

; N ≤ n] ≤

E M

N ∧n

Finally, since M

is a submartingale, E M

N ∧n

≤ E M

We now look at (b). Let us write M

∗

for max

k≤n

. We have

E [M

N ∧n

; N ≤ n] =

∞

k=0

E [M

k∧n

; N = k].

Arguing as in the proof of Theorem 4.3,

E [M

k∧n

; N = k] ≤ E [M

; N = k],

and so

E [M

N ∧n

; N ≤ n] ≤

∞

k=0

E [M

; N = k] = E [M

; N ≤ n].

The last expression is at most E [M

; M

∗

≥ λ]. If we multiply (4.1) by 2λ and integrate

over λ from 0 to ∞, we obtain

∞

2λP(M

∗

≥ λ)dλ ≤ 2

∞

E [M

: M

∗

≥ λ]

= 2E

∞

∗

≥λ)

dλ

= 2E

∗

dλ

= 2E [M

∗

Using Cauchy-Schwarz, this is bounded by

2(E M

)

1/2

(E (M

∗

)

1/2

On the other hand,

∞

2λP(M

∗

≥ λ)dλ = E

∞

2λ1

∗

≥λ)

dλ

= E

∗

2λ dλ = E (M

∗

)

We therefore have

E (M

∗

)

≤ 2(E M

)

1/2

(E (M

∗

)

1/2

Suppose E (M

∗

)

< ∞. We divide both sides by (E (M

∗

)

1/2

and square both sides.

(When E (M

∗

)

is infinite, there is a way to circumvent the division by infinity.)

The last result we want is that bounded martingales converge. (The hypothesis of

boundedness can be weakened.)

Theorem 4.5. Suppose M

is a martingale bounded in absolute value by K. That is,

| ≤ K for all n. Then lim

n→∞

exists a.s.

Proof. Since M

is bounded, it can’t tend to +∞ or −∞. The only possibility is that it

might oscillate. Let a < b be two rationals. What might go wrong is that M

might be

larger than b infinitely often and less than a infinitely often. If we show the probability of

this is 0, then taking the union over all pairs of rationals (a, b) shows that almost surely

cannot oscillate, and hence must converge.

Fix a, b and let S

= min{k : M

≤ a}, T

= min{k > S

: M

≥ b}, S

= min{k >

: M

≤ a}, and so on. Let U

= max{k : T

≤ n}. U

is called the number of

upcrossings up to time n.

We can write

2K ≥ M

− M

k=1

k+1

∧n

− M

∧n

) +

∞

k=1

∧n

− M

∧n

) + (M

∧n

− M

Now take expectations. The expectation of the first sum on the right and the last term

are zero by optional stopping. The middle term is larger than (b − a)U

, so we conclude

(b − a)E U

≤ 2K.

Let n → ∞ to see that E max

< ∞, which implies max

< ∞ a.s., which is what

we needed.

5. The one step binomial asset pricing model.

Let us begin by giving the simplest possible model of a stock and see how a European

call option should be valued in this context.

Suppose we have a single stock whose price is S

. Let d and u be two numbers with

0 < d < 1 < u. Here “d” is a mnemonic for “down” and “u” for “up.” After one time unit

the stock price will be either uS

with probability P or else dS

with probability Q, where

P + Q = 1. Instead of purchasing shares in the stock, you can also put your money in the

bank where one will earn interest at rate r. Alternatives to the bank are money market

funds or bonds; the key point is that these are considered to be risk-free.

A European call option in this context is the option to buy one share of the stock

at time 1 at price K. K is called the strike price. Let S

be the price of the stock at time

1. If S

is less than K, then the option is worthless at time 1. If S

is greater than K, you

can use the option at time 1 to buy the stock at price K, immediately turn around and

sell the stock for price S

and make a profit of S

− K. So the value of the option at time

1 is

= (S

− K)

where x

is max(x, 0). The principal question to be answered is: what is the value V

the option at time 0? In other words, how much should one pay for a European call option

with strike price K?

It is possible to buy a negative number of shares of a stock. This is equivalent to

selling shares of a stock you don’t have and is called selling short. If you sell one share

of stock short, then at time 1 you must buy one share at whatever the market price is at

that time and turn it over to the person that you sold the stock short to. Similarly you

can buy a negative number of options, that is, sell an option.

You can also deposit a negative amount of money in the bank, which is the same

as borrowing. We assume that you can borrow at the same interest rate r, not exactly a

totally realistic assumption. One way to make it seem more realistic is to assume you have

a large amount of money on deposit, and when you borrow, you simply withdraw money

from that account.

We are looking at the simplest possible model, so we are going to allow only one

time step: one makes an investment, and looks at it again one day later.

Let’s suppose the price of a European call option is V

and see what conditions

one can put on V

. Suppose you start out with V

dollars. One thing you could do is

buy one option. The other thing you could do is use the money to buy ∆

shares of

stock. If V

> ∆

, there will be some money left over and you put that in the bank. If

< ∆

, you do not have enough money to buy the stock, and you make up the shortfall

by borrowing money from the bank. In either case, at this point you have V

− ∆

the bank and ∆

shares of stock.

If the stock goes up, at time 1 you will have

∆

+ (1 + r)(V

− ∆

and if it goes down,

∆

+ (1 + r)(V

− ∆

We have not said what ∆

should be. Let us do that now. Let V

= (uS

− K)

and V

= (dS

− K)

. Note these are deterministic quantities, i.e., not random. Let

∆

− V

− dS

and we will also need

1 + r

1 + r − d

u − d

u − (1 + r)

u − d

After some algebra, we see that if the stock goes up and you had bought stock

instead of the option you would now have

+ (1 + r)(V

− W

while if the stock went down, you would now have

+ (1 + r)(V

− W

Let’s check the first of these, the second being similar. We need to show

∆

+ (1 + r)(V

− ∆

) = V

+ (1 + r)(V

− W

(5.1)

The left hand side is equal to

∆

(u − (1 + r)) + (1 + r)V

− V

u − d

(u − (1 + r)) + (1 + r)V

(5.2)

The right hand side of (5.1) is equal to

−

1 + r − d

u − d

u − (1 + r)

u − d

+ (1 + r)V

(5.3)

Now check that the coefficients of V

, of V

, and of V

agree in (5.2) and (5.3).

Suppose that V

> W

. What you want to do is come along with no money, sell

one option for V

dollars, use the money to buy ∆

shares, and put the rest in the bank

(or borrow if necessary). If the buyer of your option wants to exercise the option, you give

him one share of stock and sell the rest. If he doesn’t want to exercise the option, you sell

your shares of stock and pocket the money. Remember it is possible to have a negative

number of shares. You will have cleared (1 + r)(V

− W

), whether the stock went up or

down, with no risk.

If V

< W

, you just do the opposite: sell ∆

shares of stock short, buy one option,

and deposit or make up the shortfall from the bank. This time, you clear (1 + r)(W

− V

whether the stock goes up or down.

Now most people believe that you can’t make a profit on the stock market without

taking a risk. The name for this is “no free lunch,” or “arbitrage opportunities do not

exist.” The only way to avoid this is if V

= W

. In other words, we have shown that the

only reasonable price for the European call option is W

The “no arbitrage” condition is not just a reflection of the belief that one cannot get

something for nothing. It also represents the belief that the market is freely competitive.

The way it works is this: suppose W

= $3. Suppose you could sell options at a price

= $5; this is larger than W

and you would earn V

− W

= $2 per option without risk.

Then someone else would observe this and decide to sell the same option at a price less

than V

but larger than W

, say $4. This person would still make a profit, and customers

would go to him and ignore you because they would be getting a better deal. But then a

third person would decide to sell the option for less than your competition but more than

, say at $3.50. This would continue as long as any one would try to sell an option above

price W

We will examine this problem of pricing options in more complicated contexts, and

while doing so, it will become apparent where the formulas for ∆

and W

came from. At

this point, we want to make a few observations.

Remark 5.1. First of all, if 1 + r > u, one would never buy stock, since one can always

do better by putting money in the bank. So we may suppose 1 + r < u. We always have

1 + r ≥ 1 > d. If we set

p =

1 + r − d

u − d

q =

u − (1 + r)

u − d

then p, q ≥ 0 and p + q = 1. Thus p and q act like probabilities, but they have nothing to

do with P and Q. Note also that the price V

= W

does not depend on P or Q. It does

depend on p and q, which seems to suggest that there is an underlying probability which

controls the option price and is not the one that governs the stock price.

Remark 5.2.

There is nothing special about European call options in our argument

above. One could let V

and V

be any two values of any option, which are paid out if the

stock goes up or down, respectively. The above analysis shows we can exactly duplicate

the result of buying any option V by instead buying some shares of stock. If in some model

one can do this for any option, the market is called complete in this model.

Remark 5.3.

If we let P be the probability so that S

= uS

with probability p and

= dS

with probability q and we let E be the corresponding expectation, then some

algebra shows that

1 + r

E V

This will be generalized later.

Remark 5.4.

If one buys one share of stock at time 0, then one expects at time 1 to

have (P u + Qd)S

. One then divides by 1 + r to get the value of the stock in today’s

dollars. Suppose instead of P and Q being the probabilities of going up and down, they

were in fact p and q. One would then expect to have (pu + qd)S

and then divide by 1 + r.

Substituting the values for p and q, this reduces to S

. In other words, if p and q were

the correct probabilities, one would expect to have the same amount of money one started

with. When we get to the binomial asset pricing model with more than one step, we will

see that the generalization of this fact is that the stock price at time n is a martingale, still

with the assumption that p and q are the correct probabilities. This is a special case of

the fundamental theorem of finance: there always exists some probability, not necessarily

the one you observe, under which the stock price is a martingale.

Remark 5.5. Our model allows after one time step the possibility of the stock going up or

going down, but only these two options. What if instead there are 3 (or more) possibilities.

Suppose for example, that the stock goes up a factor u with probability P , down a factor

d with probability Q, and remains constant with probability R, where P + Q + R = 1.

The corresponding price of a European call option would be (uS

− K)

, (dS

− K)

, or

− K)

. If one could replicate this outcome by buying and selling shares of the stock,

then the “no arbitrage” rule would give the exact value of the call option in this model.

But, except in very special circumstances, one cannot do this, and the theory falls apart.

One has three equations one wants to satisfy, in terms of V

, V

, and V

. (The “c” is

a mnemonic for “constant.”) There are however only two variables, ∆

and V

at your

disposal, and most of the time three equations in two unknowns cannot be solved.

6. The multi-step binomial asset pricing model.

In this section we will obtain a formula for the pricing of options when there are n

time steps, but each time the stock can only go up by a factor u or down by a factor d.

The “Black-Scholes” formula we will obtain is already a nontrivial result that is useful.

We assume the following.

(1) Unlimited short selling of stock

(2) Unlimited borrowing

(3) No transaction costs

(4) Our buying and selling is on a small enough scale that it does not affect the market.

We need to set up the probability model. Ω will be all sequences of length n of H’s

and T ’s. S

will be a fixed number and we define S

(ω) = u

k−j

if the first k elements

of a given ω ∈ Ω has j occurrences of H and k − j occurrences of T . (What we are doing is

saying that if the j-th element of the sequence making up ω is an H, then the stock price

goes up by a factor u; if T , then down by a factor d.) F

will be the σ-field generated by

, . . . , S

Let

p =

(1 + r) − d

u − d

q =

u − (1 + r)

u − d

and define P(ω) = p

n−j

if ω has j appearances of H and n − j appearances of T . It

is not hard to see that under P the random variables S

k+1

are independent and equal

to u with probability p and d with probability q. To see this, let Y

= S

k−1

. Then

P(Y

= y

, . . . , Y

= y

) = p

n−j

, where j is the number of the y

that are equal to u. On

the other hand, this is equal to P(Y

= y

) · · · P(Y

= y

). Let E denote the expectation

corresponding to P.

The P we construct may not be the true probabilities of going up or down. That

doesn’t matter - it will turn out that using the principle of “no arbitrage,” it is P that
governs the price.

Our first result is the fundamental theorem of finance in the current context.

Proposition 6.1. Under P the discounted stock price (1 + r)

−k

is a martingale.

Proof. Since the random variable S

k+1

is independent of F

, we have

E [(1 + r)

−(k+1)

k+1

| F

] = (1 + r)

−k

(1 + r)

−1

E [S

k+1

| F

Using the independence the conditional expectation on the right is equal to

E [S

k+1

] = pu + qd = 1 + r.

Substituting yields the proposition.

Let ∆

be the number of shares held between times k and k + 1. We require ∆

to be F

measurable. ∆

, ∆

, . . . is called the portfolio process. Let W

be the amount of

money you start with and let W

be the amount of money you have at time k. W

is the

wealth process. Then

k+1

= ∆

k+1

+ (1 + r)[W

− ∆

Note that in the case where r = 0 we have

k+1

− W

= ∆

k+1

− S

k+1

i=0

∆

i+1

− S

This is a discrete version of a stochastic integral. Since

E [W

k+1

− W

| F

] = ∆

E [S

k+1

− S

| F

] = 0,

it follows that W

is a martingale. More generally

Proposition 6.2. Under P the discounted wealth process (1 + r)

−k

is a martingale.

Proof. We have

(1 + r)

−(k+1)

k+1

= (1 + r)

−k

+ ∆

[(1 + r)

−(k+1)

k+1

− (1 + r)

−k

and so

E [∆

[(1 + r)

−(k+1)

k+1

− (1 + r)

−k

| F

]

= ∆

E [(1 + r)

−(k+1)

k+1

− (1 + r)

−k

| F

] = 0.

The result follows.

Our next result is that the binomial model is complete. It is easy to lose the idea

in the algebra, so first let us try to see why the theorem is true.

For simplicity suppose r = 0. Let V

= E [V | F

]; we saw that V

is a martingale.

We want to construct a portfolio process so that W

= V . We will do it inductively by

arranging matters so that W

= V

for all k. Recall that W

is also a martingale.

Suppose we have W

= V

at time k and we want to find ∆

so that W

k+1

= V

k+1

At the (k + 1)-st step there are only two possible changes for the price of the stock and so

since V

k+1

is F

k+1

measurable, only two possible values for V

k+1

. We need to choose ∆

so that W

k+1

= V

k+1

for each of these two possibilities. We only have one parameter, ∆

to play with to match up two numbers, which may seem like an overconstrained system of

equations. But both V and W are martingales, which is why the system can be solved.

Now let us turn to the details.

Theorem 6.3. The binomial asset pricing model is complete.

Proof. Let

= (1 + r)

E [(1 + r)

−n

V | F

]

so that (1 + r)

−k

is a martingale. If ω = (t

, . . . , t

), where each t

is an H or T , let

∆

(ω) =

k+1

, . . . , t

, H, t

k+2

, . . . , t

) − V

k+1

, . . . , t

, T, t

k+2

, . . . , t

)

k+1

, . . . , t

, H, t

k+2

, . . . , t

) − S

k+1

, . . . , t

, T, t

k+2

, . . . , t

)

Set W

= V

, and we will show by induction that the wealth process at time k + 1 equals

k+1

The first thing to show is that ∆

is F

measurable. Neither S

k+1

nor V

k+1

depends

on t

k+2

, . . . , t

. So ∆

depends only on the variables t

, . . . , t

, hence is F

measurable.

Now t

k+2

, . . . , t

play no role in the rest of the proof, and t

, . . . , t

will be fixed,

so we drop the t’s from the notation.

We know (1 + r)

−k

is a martingale under P so that

= E [(1 + r)

−1

k+1

| F

]

1 + r

[pV

k+1

(H) + qV

k+1

(T )].

We now suppose W

= V

and want to show W

k+1

(H) = V

k+1

(H) and W

k+1

(T ) =

k+1

(T ). Then using induction we have W

= V

= V as required. We show the first

equality, the second being similar.

k+1

(H) = ∆

k+1

(H) + (1 + r)[W

− ∆

]

= ∆

[uS

− (1 + r)S

] + (1 + r)V

k+1

(H) − V

k+1

(T )

(u − d)S

[u − (1 + r)] + pV

k+1

(H) + qV

k+1

(T )

= V

k+1

(H).

We are done.

Finally, we obtain the Black-Scholes formula in this context. Let V be any option

that is F

-measurable. The one we have in mind is the European call, for which V =

− K)

, but the argument is the same for any option whatsoever.

Theorem 6.4. The value of the option V at time 0 is V

= (1 + r)

−n

E V .

Proof. We can construct a portfolio process ∆

so that if we start with W

= (1+r)

−n

E V ,

then the wealth at time n will equal V , no matter what the market does in between. If

we could buy or sell the option V at a price other than W

, we could obtain a riskless

profit. That is, if the option V could be sold at a price c

larger than W

, we would sell

the option for c

dollars, use W

to buy and sell stock according to the portfolio process

∆

, have a net worth of V + (1 + r)

− W

) at time n, meet our obligation to the buyer

of the option by using V dollars, and have a net profit, at no risk, of (1 + r)

− W

If c

were less than W

, we would do the same except buy an option, hold −∆

shares at

time k, and again make a riskless profit. By the “no arbitrage” rule, that can’t happen,

so the price of the option V must be W

Remark 6.5.

Note that the proof of Theorem 6.4 tells you precisely what hedging

strategy (i.e., what portfolio process to use).

In the binomial asset pricing model, there is no difficulty computing the price of a

European call. We have

E (S

− K)

(x − K)

P(S

= x)

and

P(S

= x) =

n−k

if x = u

n−k

. Therefore the price of the European call is

k=0

n−k

− K)

n−k

The formula in Theorem 6.4 holds for exotic options as well. Suppose

V = max

i=1,...,n

− min

j=1,...,n

In other words, you sell the stock for the maximum value it takes during the first n time

steps and you buy at the minimum value the stock takes; you are allowed to wait until

time n and look back to see what the maximum and minimum were. You can even do this

if the maximum comes before the minimum. This V is still F

measurable, so the theory

applies. Naturally, such a “buy low, sell high” option is very desirable, and the price of

such a V will be quite high. It is interesting that even without using options, you can

duplicate the operation of buying low and selling high by holding an appropriate number

of shares ∆

at time k, where you do not look into the future to determine ∆

7. American options.

An American option is one where you can exercise the option any time before some

fixed time T . For example, on a European call, one can only use it to buy a share of stock

at the expiration time T , while for an American call, at any time before time T , one can

decide to pay K dollars and obtain a share of stock.

Let us give an informal argument on how to price an American call, giving a more

rigorous argument in a moment. One can always wait until time T to exercise an American

call, so the value must be at least as great as that of a European call. On the other hand,

suppose you decide to exercise early. You pay K dollars, receive one share of stock, and

your wealth is S

− K. You hold onto the stock, and at time T you have one share of stock

worth S

, and for which you paid K dollars. So your wealth is S

− K ≤ (S

− K)

. In

fact, we have strict inequality, because you lost the interest on your K dollars that you

would have received if you had waited to exercise until time T . Therefore an American

call is worth no more than a European call, and hence its value must be the same as that

of a European call.

This argument does not work for puts, because selling stock gives you some money

on which you will receive interest, so it may be advantageous to exercise early. (A put is

the option to sell a stock at a price K at time T .)

Here is the more rigorous argument. Let g(x) be convex with g(0) = 0. Certainly

g(x) = (x − K)

is such a function. We have

g(λx) = g(λx + (1 − λ) · 0) ≤ λg(x) + (1 − λ)g(0) = λg(x).

By Jensen’s inequality,

E [(1 + r)

−(k+1)

g(S

k+1

) | F

] = (1 + r)

−k

1 + r

g(S

k+1

) | F

≥ (1 + r)

−k

1 + r

k+1

| F

≥ (1 + r)

−k

1 + r

k+1

| F

= (1 + r)

−k

g(S

So (1 + r)

−k

g(S

) is a submartingale. By optional stopping,

E [(1 + r)

−τ

g(S

)] ≤ E [(1 + r)

−n

g(S

)],

so τ ≡ n always does best.

8. Continuous random variables.

We are now going to start working towards continuous times and stocks that can

take any positive number as a value, so we need to prepare by extending some of our

definitions.

Given any random variable X, we can approximate it by r.v’s X

that are discrete.

We let

i=−n2

(i/2

≤X<(i+1)/2

)

In words, if X(ω) lies between −n and n, we let X

(ω) be the closest value i/2

that

is less than or equal to X(ω). For ω where |X(ω)| > n we set X

(ω) = 0. Clearly the

are discrete, and approximate X. In fact, on the set where |X| ≤ n, we have that

|X(ω) − X

(ω)| ≤ 2

−n

For reasonable X we are going to define E X = lim E X

. There are some things

one wants to prove, but all this has been worked out in measure theory and the theory of

the Lebesgue integral. Let us confine ourselves here to showing this definition is the same

as the usual one when X has a density.

Recall X has a density f

P(X ∈ [a, b]) =

(x)dx

for all a and b. In this case

E X =

∞

−∞

(x)dx.

With our definition of X

we have

P(X

= i/2

) = P(X ∈ [i/2

, (i + 1)/2

)) =

(i+1)/2

i/2

(x)dx.

Then

E X

P(X

= i/2

) =

(i+1)/2

i/2

(x)dx.

Since x differs from i/2

by at most 1/2

when x ∈ [i/2

, (i + 1)/2

), this will tend to

R xf

(x)dx, unless the contribution to the integral for |x| ≥ n does not go to 0 as n → ∞.

As long as

R |x|f

(x)dx < ∞, one can show that this contribution does indeed go to 0.

We also need an extension of the definition of conditional probability. A r.v. is G

measurable if (X > a) ∈ G for every a. How do we define E [Z | G] when G is not generated
by a countable collection of disjoint sets?

Again, there is a completely worked out theory that holds in all cases. Let us give

a definition that is equivalent that works except for a very few cases. Let us suppose that

for each n the σ-field G

is finitely generated. This means that G

is generated by finitely

many disjoint sets B

, . . . , B

. So for each n, the number of B

is finite but arbitrary,

the B

are disjoint, and their union is Ω. Suppose also that G

⊂ G

⊂ · · ·. Now ∪

will not in general be a σ-field, but suppose G is the smallest σ-field that contains all the

. Finally, define P(A | G) = lim P(A | G

This is a fairly general set-up. For example, let Ω be the real line and let G

generated by the sets (−∞, n), [n, ∞) and [i/2

, (i + 1)/2

). Then G will contain every

interval that is closed on the left and open on the right, hence G must be the σ-field that

one works with when one talks about Lebesgue measure on the line.

The question that one might ask is: how does one know the limit exists? Since the

increase, we know that M

= P(A | G

) is a martingale with respect to the G

. It is

certainly bounded above by 1 and bounded below by 0, so by the martingale convergence

theorem, it must have a limit as n → ∞.

Once one has a definition of conditional probability, one defines conditional expec-

tation by what one expects. If X is discrete, one can write X as

and then one

defines

E [X | G] =

P(A

| G

If the X is not discrete, one approximates as above. One has to worry about convergence,

but everything does go through.

With this extended definition of conditional expectation, do all the properties of

Section 2 hold? The answer is yes, and the proofs are by taking limits of the discrete

approximations.

We will be talking about stochastic processes. Previously we discussed sequences

, S

, . . . of r.v.’s. Now we want to talk about processes Y

for t ≥ 0. We typically let

be the smallest σ-field with respect to which Y

is measurable for all s ≤ t. As you

might imagine, there are a few technicalities one has to worry about. We will try to avoid

thinking about them as much as possible.

A continuous time martingale (or submartingale) is what one expects: M

is in-

tegrable, adapted to F

, and if s < t, then E [M

| F

] = M

. The analogues of Doob’s

theorems go through. The way to prove these is to observe that M

k/2

is a discrete time

martingale, and then to take limits as n → ∞.

9. Brownian motion.

Let S

be a simple symmetric random walk. This means that Y

= S

− S

k−1

equals +1 with probability

1
2

, equals −1 with probability

1
2

, and is independent of Y

for

j < k. We notice that E S

= 0 while E S

n
i=1

E Y

i6=j

E Y

= n using the fact

that E [Y

] = (E Y

)(E Y

) = 0.

Define X

= S

√

n if nt is an integer and by linear interpolation for other t. If

nt is an integer, E X

= 0 and E (X

)

= t. It turns out X

does not converge for any ω.

However there is another kind of convergence, called weak convergence, that takes

place. There exists a process Z

such that for each k, each t

< t

< · · · < t

, and each

< b

, a

< b

, . . . , a

< b

, we have

(1) The paths of Z

are continuous as a function of t.

(2) P(X

∈ [a

, b

], . . . , X

∈ [a

, b

]) → P(Z

∈ [a

, b

], . . . , Z

∈ [a

, b

]).

The limit Z

is called a Brownian motion starting at 0. It has the following properties.

(1) E Z

= 0.

(2) E Z

= t.

(3) Z

− Z

is independent of F

= σ(Z

, r ≤ s).

(4) Z

− Z

has the distribution of a normal random variable with mean 0 and variance

t − s. This means

P(Z

− Z

∈ [a, b]) =

p2π(t − s)

−y

/2(t−s)

dy.

(This result follows from the central limit theorem.)

(5) The map t → Z

(ω) is continuous for each ω.

10. Markov properties of Brownian motion.

Fix r and let W

= Z

t+r

− Z

. Clearly the map t → W

is continuous since the

same is true for Z. Since W

− W

= Z

t+r

− Z

s+r

, then the distribution of W

− W

normal with mean zero and variance (t + r) − (s + r). One can also check the other parts

of the definition and we see that W

is also a Brownian motion. This is a version of the

Markov property. We will prove the following stronger result, which is a version of the

strong Markov property.

A stopping time in the continuous framework is a r.v. T taking values in [0, ∞)

such that (T > t) ∈ F

for all t. To make a satisfactory theory, one needs that the F

what is called right continuous: F

= ∩

ε>0

t+ε

, but this is fairly technical and we will

ignore it.

If T is a stopping time, F

is the collection of events A such that A ∩ (T > t) ∈ F

for all t.

Proposition 10.1. If X

is a Brownian motion and T is a bounded stopping time, then

T +t

− X

is a mean 0 variance t random variable and is independent of F

Proof. Let T

be defined by T

(ω) = (k + 1)/2

if T (ω) ∈ [k/2

, (k + 1)/2

). It is easy

to check that T

is a stopping time. Let f be continuous and A ∈ F

. Then A ∈ F

well. We have

E [f (X

− X

); A] =

E [f (X

− X

); A ∩ T

= k/2

]

E [f (X

− X

)]P(A ∩ T

= k/2

)

= E f (X

)P(A).

Let n → ∞, so

E [f (X

T +t

− X

); A] = E f (X

)P(A).

Taking limits this equation holds for all bounded f .

If we take A = Ω and f = 1

, we see that X

T +t

− X

has the same distribution

as X

, which is that of a mean 0 variance t normal random variable. If we let A ∈ F

arbitrary and f = 1

, we see that

P(X

T +t

− X

∈ B, A) = P(X

∈ B)P(A) = P(X

T +t

− X

∈ B)P(A),

which implies that X

T +t

− X

is independent of F

This proposition says: if you want to predict X

T +t

, you could do it knowing all of

or just knowing X

. Since X

T +t

− X

is independent of F

, the extra information

given in F

does you no good at all.

We need a way of expressing the Markov and strong Markov properties that will

generalize to other processes.

Let W

be a Brownian motion. Consider the process W

= x+W

, Brownian motion

started at x. Define Ω

to be set of continuous functions on [0, ∞), let X

(ω) = ω(t), and

let the σ-field be the one generated by the X

. Define P

on (Ω

, F

) by

∈ A

, . . . , X

∈ A

) = P(W

∈ A

, . . . , W

∈ A

What we have done is gone from one probability space Ω with many processes W

to one

process X

with many probability measures P

Proposition 10.2. If s < t and f is bounded or nonnegative, then

[f (X

) | F

] = E

[f (X

t−s

)],

a.s.

The right hand side is to be interpreted as follows. Define ϕ(x) = E

f (X

t−s

). Then

f (X

t−s

) means ϕ(X

(ω)). One often writes P

f (x) for E

f (X

Before proving this, recall from undergraduate analysis that every bounded function

is the limit of linear combinations of functions e

iux

, u ∈ R. This follows from using the

inversion formula for Fourier transforms. There are various slightly different formulas for

the Fourier transform. We use b

f (u) =

R e

iux

f (x) dx. If f is smooth enough and has

compact support, then one can recover f by the formula f (x) =

2π

R e

−iux

f (u) du. We

can first approximate this improper integral by

2π

−N

−iux

f (u) du by taking N large

enough, and then approximate the approximation by using Riemann sums. Thus we can

approximate f (x) by a linear combination of terms of the form e

. Also, bounded

functions can be approximated by smooth functions with compact support.

Proof. Let f (x) = e

iux

. Then

iuX

| F

] = e

iuX

E [e

iu(X

−X

)

| F

]

= e

iuX

−u

(t−s)/2

On the other hand,

ϕ(y) = E

[f (X

t−s

)] = E [e

iu(W

t−s

+y)

] = e

iuy

−u

(t−s)/2

So ϕ(X

) = E

iuX

| F

]. Using linearity and taking limits, we have the lemma for all

f .

This formula generalizes: If s < t < u, then

[f (X

)g(X

) | F

] = E

[f (X

t−s

)g(X

u−s

)],

and so on for functions of X at more times.

Using Proposition 10.1, the statement and proof of Proposition 10.2 can be extended

to stopping times.

Proposition 10.3. If T is a bounded stopping time, then

[f (X

T +t

) | F

] = E

[f (X

)].

11. Stochastic integrals.

If one wants to consider the (deterministic) integral

f (s) dg(s), where f and g

are continuous and g is differentiable, we can define it analogously to the usual Riemann

integral as the limit of Riemann sums

n
i=1

f (s

)[g(s

)−g(s

i−1

)], where s

< s

< · · · < s

is a partition of [0, t]. This is known as the Riemann-Stieltjes integral. One can show (using

the mean value theorem, for example) that

f (s) dg(s) =

f (s)g

(s) ds.

If we were to take f (s) = 1

[0,a]

(s), one would expect the following:

[0,a]

(s) dg(s) =

[0,a]

(s)g

(s) ds =

(s) ds = g(a) − g(0).

Note that although we use the fact that g is differentiable in the intermediate stages, the

first and last terms make sense for any g.

We now want to replace g by a Brownian path and f by a random integrand. The

expression

R f (s) dW (s) does not make sense as a Riemann-Stieltjes integral because it is

a fact that W (s) is not differentiable as a function of t. We need to define the expression

by some other means. We will show that it can be defined as the limit in L

of Riemann

sums. The resulting integral is called a stochastic integral.

Let us consider a very special case first. Suppose f is continuous and determin-

istic (i.e., does not depend on ω).

Suppose we take a Riemann sum approximation

P f (

)[W (

i+1

) − W (

)]. If we take the difference of two successive approximations

we have terms like

i odd

[f (i/2

n+1

) − f ((i + 1)/2

n+1

)][W ((i + 1)/2

n+1

) − W (i/2

n+1

)].

This has mean zero. By the independence, the second moment is

[f (i/2

n+1

) − f ((i + 1)/2

n+1

)]

(1/2

n+1

This will be small if f is continuous. So by taking a limit in L

we obtain a nontrivial

limit.

We now turn to the general case. Let W

be a Brownian motion. We will only

consider integrands H

such that H

is F

measurable for each s.

We will construct

for all H with E

ds < ∞.

Before we proceed we will need to define the quadratic variation of a continuous

martingale. We will use the following theorem without proof because in our applications

we can construct the desired increasing process directly.

Theorem 11.1. Suppose M

is a continuous martingale such that E M

< ∞ for all t.

There exists one and only one increasing process A

that is adapted to F

, has continuous

paths, and A

= 0 such that M

− A

is a martingale.

The simplest example of such a martingale is Brownian motion. If W

is a Brownian

motion, let L

= W

− t. Note

E [L

| F

] = E [((W

− W

) + W

)

| F

] − t

= E [(W

− W

)

| F

] − 2E [W

− W

) | F

] + E [W

| F

] − t.

The first term, using independence, is E [(W

−W

)

] = t−s, the second term is W

E [W

−

| F

] = W

E [W

− W

] = 0, and so we have

E [L

| F

] = (t − s) + W

− t = L

or L

is a martingale. So in the case of Brownian motion, the increasing process A

= t.

We use the notation hM i

for the increasing process given in Theorem 11.1. We

will see that in the case of stochastic integrals, where N

, it turns out that

hN i

ds.

If K is bounded and F

measurable, let N

= K(W

t∧b

−W

t∧a

). Part of the statement

of the next proposition is that hN i

exists.

Lemma 11.2. N

is a continuous martingale, E N

∞

= E [K

(b − a)] and

hN i

[a,b]

(s) ds.

Proof. The continuity is clear. Let us look at E [N

| F

]. In the case a < s < t < b, this

is equal to

E [K(W

− W

) | F

] = KE [(W

− W

) | F

] = K(W

− W

) = N

In the case s < a < t < b, E [N

| F

] is equal to

E [K(W

− W

) | F

] = E [KE [W

− W

| F

] | F

] = 0 = N

The other possibilities for where s and t are can be done similarly.

Recall W

− t is a martingale. For E N

∞

, we have

E N

∞

= E [K

E [(W

− W

)

| F

]] = E [K

E [W

− W

| F

]]

= E [K

E [b − a | F

]] = E [K

(b − a)].

For hN i

, we need to show

E [K

t∧b

− W

t∧a

)

− K

(t ∧ b − t ∧ a) | F

]

= K

s∧b

− W

s∧a

)

− K

(s ∧ b − s ∧ a).

We do this by checking all the cases.

is said to be simple if it can be written in the form

J
j=1

]

(s), where

is F

measurable and bounded. Define

j=1

∧t

− W

∧t

Proposition 11.3. N

is a continuous martingale, E N

∞

= E

∞

ds, and hN i

ds.

Proof. We may rewrite H so that the intervals [a

, b

] satisfy a

≤ b

≤ a

≤ b

≤ · · · ≤ b

It is then clear that N

is a martingale.

We have

E N

∞

= E

h X

− W

)

+ 2E

h X

i<j

− W

)(W

− W

)

The cross terms vanish, because when we condition on F

, we have

E [H

− W

)E [(W

− W

) | F

] = 0.

For the diagonal terms

E [H

− W

)

] = E [H

E [(W

− W

)

| F

]]

= E [H

E [W

− W

| F

]]

= E [H

E [b

− a

| F

]]

= E [H

([b

− a

)].

So E N

∞

= E

∞

ds.

Now suppose H

is adapted and E

∞

ds < ∞. Let us define some norms; these

are just for technical purposes and can be ignored if wished. Let

∞

1/2

This is the L

norm on [0, ∞) × Ω with respect to the measure ν

(A) = E

∞

ds. Define

another norm by

kY k

= (E [sup

])

1/2

Using some results from measure theory, we can choose H

simple such that E

∞

−

)

ds → 0. (This is the fact that simple functions are dense in L

.) By Doob’s inequality

we have

sup

− H

) dW

≤ 4E

∞

− H

) dW

= 4E

∞

− H

)

ds → 0.

One can show that the norm kY k

is complete, so there exists a process N

such that

sup

[

− N

] → 0 in L

If H

and H

are two sequences converging to H, then E (

− H

) dW

)

− H

)

ds → 0, or the limit is independent of which sequence H

we choose. It

is easy to see, because of the L

convergence, that N

is a martingale, E N

= E

ds,

and hN i

ds. Because sup

[

− N

] → 0 in L

, one can show there exists

a subsequence such that the convergence takes place almost surely. So with probability

one, N

has continuous paths. We write N

and call N

the stochastic integral

of H with respect to W .

We discuss some extensions of the definition. First of all, if we replace W

by a

continuous martingale M

and H

is adapted with E

dhM i

< ∞, we can dupli-

cate everything we just did with ds replaced by dhM i

and get a stochastic integral. In

particular, if dhM i

= K

ds, we replace ds by K

ds.

There are some other extensions of the definition that are not hard. If

∞

hM i

∞ but without the expectation being finite, we can define the stochastic integral by looking

at M

t∧T

for suitable stopping times T

and then letting T

→ ∞.

A process A

is of bounded variation if the paths of A

have bounded variation. This

means that one can write A

= A

+
t

−A

−
t

, where A

+
t

and A

−
t

have paths that are increasing.

|A|

is then defined to be A

+
t

+ A

−
t

. A semimartingale is the sum of a martingale and a

process of bounded variation. If

∞

dhM i

∞

| |dA

| < ∞ and X

= M

+ A

we define

where the first integral on the right is a stochastic integral and the second is a Riemann-

Stieltjes or Lebesgue-Stieltjes integral. For a semimartingale, we define hXi

= hM

Given two semimartingales X and Y we define hX, Y i

by what is known as polar-

ization:

hX, Y i

1
2

[hX + Y i

− hXi

− hY i

As an example, if X

and Y

, then (X + Y )

+ K

)dW

hX + Y i

+ K

)

ds =

ds +

ds.

Since hXi

ds with a similar formula for hY i

, we conclude

hX, Y i

ds.

What does a stochastic integral mean? If one thinks of the derivative of Z

as being

a white noise, then

is like a filter that increases or decreases the volume by a

factor H

For us, an interpretation is that Z

represents a stock price. Then

repre-

sents our profit (or loss) if we hold H

shares at time s. This can be seen most easily if

= G1

[a,b]

. So we buy G(ω) shares at time a and sell them at time b. The stochastic

integral represents our profit or loss.

Since we are in continuous time, we are allowed to buy and sell continuously and

instantaneously.

What we are not allowed to do is look into the future to make our

decisions, which is where the H

adapted condition comes in.

12. Ito’s formula.

Suppose W

is a Brownian motion and f : R → R is a C

function, that is, f and its

first two derivatives are continuous. Ito’s formula, which is sometime known as the change

of variables formula, says that

f (W

) − f (W

) =

)ds +

1
2

)ds.

Compare this with the fundamental theorem of calculus:

f (t) − f (0) =

(s)ds.

In Ito’s formula we have a second order term to carry along.

The idea behind the proof is quite simple. By Taylor’s theorem.

f (W

) − f (W

) =

n−1

i=0

[f (W

(i+1)t/n

) − f (W

it/n

)]

≈

n−1

i=1

it/n

)(W

(i+1)t/n

− W

it/n

)

1
2

n−1

i=0

it/n

)(W

(i+1)t/n

− W

it/n

)

The first sum on the right is approximately the stochastic integral and the second is

approximately the quadratic variation.

For a more general semimartingale X

= M

+ A

, Ito’s formula reads

Theorem 12.1. If f ∈ C

, then

f (X

) − f (X

) =

)dXs +

1
2

)dhM i

Let us look at an example. Let W

be Brownian motion, X

= σW

− σ

t/2, and

f (x) = e

. Then hXi

= hσW i

= σ

t and

σW

−σ

t/2

= 1 +

σdW

−

1
2

(12.1)

1
2

= 1 +

σdW

For a semimartingale X

= M

we set hXi

= hM i

. Given two semimartingales

X, Y , we define

hX, Y i

1
2

[hX + Y i

− hXi

− hY i

Proposition 12.2.

= X

+ hX, Y i

Proof. Applying Ito’s formula with f (x) = x

to X

+ Y

, we obtain

+ Y

)

= (X

+ Y

)

+ 2

+ Y

)(dX

+ dY

) + hX + Y i

Applying Ito’s formula with f (x) = x

to X and to Y , then

= X

+ 2

+ hXi

and

= Y

+ 2

+ hY i

Then some algebra and the fact that

1
2

[(X

+ Y

)

− X

− Y

]

yields the formula.

There is a multidimensional version of Ito’s formula: if X

= (X

, . . . , X

) is a

vector, each component of which is a semimartingale, and f ∈ C

, then

f (X

) − f (X

) =

i=1

∂f

∂x

)dX

) +

1
2

i,j=1

∂

∂x

2
i

)dhX

, X

The following application of Ito’s formula, known as L´

evy’s theorem, is important.

Theorem 12.3. Suppose M

is a continuous martingale with hM i

= t. Then M

is a

Brownian motion.

Before proving this, recall from undergraduate probability that the moment generating

function of a r.v. X is defined by m

(a) = E e

and that if two random variables have

the same moment generating function, they have the same law. This is also true if we

replace a by iu. In this case we have ϕ

(u) = E e

iuX

and ϕ

is called the characteristic

function of X. The reason for looking at the characteristic function is that ϕ

always

exists, whereas m

(a) might be infinite. The one special case we will need is that if X is

a normal r.v. with mean 0 and variance t, then ϕ

(u) = e

−u

t/2

. This follows from the

formula for m

(a) with a replaced by iu (this can be justified rigorously).

Proof. Apply Ito’s formula with f (x) = e

iux

. Then

iuM

= 1 +

iue

iuM

1
2

(−u

iuM

dhM i

Taking expectations and using hM i

= s and the fact that a stochastic integral is a

martingale, hence has 0 expectation, we have

E e

iuM

= 1 −

iuM

ds.

Let J (t) = E e

iuM

. The equation can be rewritten

J (t) = 1 −

J (s)ds.

So J

(t) = −

1
2

J (t) with J (0) = 1. The solution to this elementary ODE is J (t) =

−u

t/2

, which shows that M

is a mean 0 variance t normal r.v.

If A ∈ F

and we do the same argument with M

replaced by M

s+t

− M

, we have

iu(M

s+t

−M

)

= 1 +

iue

iu(M

s+r

−M

)

1
2

(−u

iu(M

s+r

−M

)

dhM i

Multiply this by 1

and take expectations. Since a stochastic integral is a martingale, the

stochastic integral term again has expectation 0. If we let K(t) = E [e

iu(M

t+s

−M

)

; A], we

now arrive at K

(t) = −

1
2

K(t) with K(0) = P(A), so

K(t) = P(A)e

−u

t/2

Therefore

iu(M

t+s

−M

)

; A

= E e

iu(M

t+s

−M

)

P(A).

(12.2)

If f is a nice function and b

f is its Fourier transform, replace u in the above by −u,

multiply by b

f (u), and integrate over u. (To do the integral, we approximate the integral

by a Riemann sum and then take limits.) We then have

E [f (M

s+t

− M

); A] = E [f ((M

s+t

− M

)]P(A).

By taking limits we have this for f = 1

, so

P(M

s+t

− M

∈ B, A) = P(M

s+t

− M

∈ B)P(A).

This implies that M

s+t

− M

is independent of F

Note Var (M

− M

) = t − s; take A = Ω in (12.2).

13. The Girsanov theorem.

Suppose P is a probability measure and

= dW

+ µ(X

)dt.

Let

= exp

−

µ(X

)dW

−

µ(X

)

ds/2

Then as we have seen before, by Ito’s formula, M

is a martingale.

Now let us define a new probability by setting

Q(A) = E [M

; A]

(13.1)

if A ∈ F

. We had better be sure this Q is well defined. If A ∈ F

⊂ F

, then E [M

; A] =

E [M

; A] because M

is a martingale.

What the Girsanov theorem says is

Theorem 13.1. Under Q, X

is a Brownian motion.

There is a more general version.

Theorem 13.2. If X

is a martingale under P, then under Q the process X

− D

is a

martingale where

dhX, M i

hXi

is the same under both P and Q.

Let us see how Theorem 13.1 can be used. Let S

be the stock price, and suppose

= σS

+ µS

dt.

Define

= e

(−µ/σ)(W

)−(µ

/2σ

Then from (12.1) M

is a martingale and

= 1 +

−

Let X

= W

. Then

hX, M i

−

ds = −

ds.

Therefore

dhX, M i

= −

ds = −(µ/σ)t.

Define Q by (13.1). By Theorem 13.2, under Q the process f

= W

+ (µ/σ)t is a

martingale. Hence

= σS

(dW

+ (µ/σ)dt) = σS

= S

σS

is a martingale. So we have found a probability under which the asset price is a martingale.

This means that Q is the risk-neutral probability, which we have been calling P.

Let us give another example of the use of the Girsanov theorem. Suppose X

+ µt. We want to compute the probability that X

exceeds the level a by time t

We first need the probability that a Brownian motion crosses a level a by time

. Any path that crosses a but is at level x < a at time t

has a corresponding path

determined by reflecting across level a at the first time the Brownian motion hits a; the

reflected path will end up at a + (a − x) = 2a − x. This is known as the reflection principle,

and can be written, informally, by

P( sup

s≤t

≥ a, W

= x) = P(W

= 2a − x).

Now let W

be a Brownian motion under P. Let dQ/dP = M

= e

µW

−µ

t/2

. Let

= W

− µt. Theorem 13.1 says that under Q, Y

is a Brownian motion. We have

= Y

+ µt.

Let A = (sup

s≤t

≥ a). We want to calculate

P( sup

s≤t

+ µs) ≥ a).

is a Brownian motion under P while Y

is a Brownian motion under Q. So this proba-

bility is equal to

Q( sup

s≤t

+ µs) ≥ a).

This in turn is equal to

Q( sup

s≤t

≥ a) = Q(A).

Now we use the expression for M

Q(A) = E

µW

−µ

; A]

∞

−∞

µx−µ

P( sup

s≤t

≥ a, W

= x)dx

= e

−µ

−∞

√

2πt

µx

−(2a−x)

/2t

dx +

∞

√

2πt

µx

−x

/2t

dx.

Now for the proofs of Theorems 13.1 and 13.2.

Proof of Theorem 13.2. Assume without loss of generality that X

= 0. Then if A ∈ F

; A] = E

; A]

= E

; A

+ E

; A

+ E

[hX, M i

; A]

= E

; A

+ E

; A

+ E

[hX, M i

; A]

= E

; A] + E [hX, M i

− hX, M i

; A].

Here we used the fact that stochastic integrals with respect to the martingales X and M

are again martingales.

On the other hand,

[hX, M i

− hX, M i

; A] = E

dhX, M i

; A

= E

; A

= E

| F

] dD

; A

= E

; A

= E

[(D

− D

; A]

= E

− D

; A].

The quadratic variation proof is similar.

Proof of Theorem 13.1. From our formula for M we have dM

= −M

µ(X

)dW

, so

dhX, M i

= −M

µ(X

)dt. Hence by Theorem 13.2 we see that under Q, X

is a continuous

martingale with hXi

= t. By L´

evy’s theorem, this means that X is a Brownian motion

under Q.

To help understand what is going on, let us give another proof of Theorem 13.1

along the lines of the proof of Theorem 13.2.

Proof of Theorem 13.1, second version. Using Ito’s formula with f (x) = e

= 1 −

µ(X

hW, M i

= −

µ(X

dr.

Since Q(A) = E

; A], it is not hard to see that

; A] = E

; A].

By Ito’s product formula this is

; A

+ E

; A

+ E

hW, M i

; A

Since

and

are stochastic integrals with respect to martingales, they

are themselves martingales. Thus the above is equal to

; A

+ E

; A

+ E

hW, M i

; A

Using the product formula again, this is

; A] + E

[hW, M i

− hW, M i

; A] = E

; A] + E

[hW, M i

− hW, M i

; A].

The last term on the right is equal to

dhW, M i

; A

= E

−

µ(X

)dr; A

= E

−

E [M

| F

]µ(X

)dr; A

= E

−

µ(X

)dr; A

= E

−

µ(X

)dr; A

= −E

µ(X

) dr; A

+ E

µ(X

) dr; A

Therefore

µ(X

)dr; A

= E

µ(X

)dr; A

which shows X

is a martingale with respect to Q.

Similarly, X

− t is a martingale with respect to Q. By L´evy’s theorem, X

is a

Brownian motion.

14. Stochastic differential equations.

Let W

be a Brownian motion. We are interested in the existence and uniqueness

for stochastic differential equations (SDEs) of the form

= σ(X

) dW

+ b(X

) dt,

= 0.

This means X

satisfies

= x

σ(X

) dW

b(X

) ds.

(14.1)

Here W

is a Brownian motion, and (14.1) holds for almost every ω.

We have to make some assumptions on σ and b. We assume they are Lipschitz,

which means:

|σ(x) − σ(y)| ≤ c|x − y|,

|b(x) − b(y)| ≤ c|x − y|

for some constant c. We also suppose that σ and b grow at most linearly, which means:

|σ(x)| ≤ c(1 + |x|),

|b(x)| ≤ c(1 + |x|).

Theorem 14.1. There exists one and only one solution to (14.1).

The idea of the proof is Picard iteration, which is how existence and uniqueness

for ordinary differential equations is proved. Let us illustrate the uniqueness part, and for

simplicity, assume b is identically 0.

Proof of uniqueness. If X and Y are two solutions,

− Y

[σ(X

) − σ(Y

)]dW

E |X

− Y

= E

|σ(X

) − σ(Y

ds ≤ c

E |X

− Y

ds,

using the Lipschitz hypothesis on σ. If we let g(t) = E |X

− Y

, we have

g(t) ≤ c

g(s) ds.

Then

g(t) ≤ c

g(r) dr

ds.

g is easily seen to be bounded on finite intervals, and iteration implies

g(t) ≤ At

/n!

for each n, which implies g must be 0.

The above theorem also works in higher dimensions. We want to solve

j=1

)dW

+ b

)ds,

i = 1, . . . , d.

If all of the σ

and b

are Lipschitz and grow at most linearly, we have uniqueness for the

solution.

Suppose one wants to solve

= aZ

+ bZ

dt.

Note that this equation is linear in Z

, and it turns out that linear equations are almost

the only ones that have an explicit solution. In this case we can write down the explicit

solution and then verify that it satisfies the SDE. The uniqueness result shows that we

have in fact found the solution.

Let

= Z

−a

t/2+bt

We will verify that this is correct by using Ito’s formula. Let X

= aW

− a

t/2 + bt. Then

is a semimartingale with martingale part aW

and hXi

= a

t. Z

= e

. By Ito’s

formula with f (x) = e

= Z

1
2

= Z

−

ds +

b ds

1
2

ds.

This is the integrated form of the equation we wanted to solve.

If we let X

denote the solution to

= x +

σ(X

)dW

b(X

)ds,

so that X

is the solution of the SDE started at x, we can define new probabilities by

∈ A

, . . . , X

∈ A

) = P(X

∈ A

, . . . , X

∈ A

This is similar to what we did in defining P

for Brownian motion, but here we do not have

translation invariance. One can show that when there is uniqueness, the family (P

, X

)

satisfies the strong Markov property.

15. Continuous time financial models.

The most common model by far in finance is that the security price is based on a

Brownian motion. One does not want to say the price is some multiple of Brownian motion

for two reasons. First, of all, a Brownian motion can become negative, which doesn’t make

sense for stock prices. Second, if one invests $1,000 in a stock selling for $1 and it goes

up to $2, one has the same profit as if one invests $1,000 in a stock selling for $100 and it

goes up to $200. It is the proportional increase one wants.

Therefore one sets ∆S

to be the quantity related to a Brownian motion. Differ-

ent stocks have different volatilities σ (consider a high-tech stock versus a pharmaceutical).

In addition, one expects a mean rate of return µ on ones investment that is positive (oth-

erwise, why not just put the money in the bank?). In fact, one expects the mean rate

of return to be higher than the risk-free interest rate r because one expects something in

return for undertaking risk.

So the model that is used is to let the stock price be modeled by the SDE

= σdW

+ µdt,

or what looks better,

= σS

+ µS

dt.

(15.1)

Fortunately this SDE is one of those that can be solved explicitly.

Proposition 15.1. The solution to (15.1) is given by

= S

σW

+(µ−(σ

/2)t)

(15.2)

Proof. Using Theorem 14.1 there will only be one solution, so we need to verify that S

as given in (15.2) satisfies (15.1). We already did this, but it is important enough that we

will do it again. Let us first assume S

= 1. Let X

= σW

+ (µ − (σ

/2)t, let f (x) = e

and apply Ito’s formula. We obtain

= e

1
2

dhXi

= 1 +

σdW

(µ −

1
2

)ds

1
2

= 1 +

σdW

µds,

which is (15.1). If S

6= 0, just multiply both sides by S

If one purchases ∆

shares (possibly a negative number) at time t

, then changes

the investment to ∆

shares at time t

, etc., then ones wealth at time t will be

+ ∆

− S

) + ∆

− S

) + · · · + ∆

i+1

− S

But this is the same as

∆(s)dS

where we have t ≥ t

i+1

and ∆(s) = ∆

if t

≤ s < t

i+1

. In other words, our wealth is

given by a stochastic integral with respect to the stock price. The requirement that the

integrand of a stochastic integral be adapted is very natural: we cannot base the number

of shares we own at time s on information that will not be available until the future.

The continuous time model of finance is then that the security price is given by

(15.1) (often called geometric Brownian motion), that there are no transaction costs, but

one can trade as many shares as one wants and vary the amount held in a continuous

fashion. This clearly is not the way the market actually works, for example, stock prices

are discrete, but this model has proved to be a very good one.

16. Martingale representation theorem.

In this section we want to show that every random variable that is F

measurable

can be written as a stochastic integral of Brownian motion. In the next section we use

this to show that under the model of geometric Brownian motion the market is complete.

This means that no matter what option one comes up with, one can exactly replicate the

result (no matter what the market does) by buying and selling shares of stock.

In mathematical terms, we let F

be the σ-field generated by W

, s ≤ t. From (15.2)

we see that F

is also the same as the σ-field generated by S

, s ≤ t, so it doesn’t matter

which one we work with. We want to show that if V is F

measurable, then there exists

adapted such that

V = V

(16.1)

where V

is a constant.

We first need the following.

Proposition 16.1. Suppose

= V

and

E |V

− V

→ 0

for each t with the H

adapted. Then there exists H

adapted so that

= V

What this proposition says is that if we can duplicate a sequence of options V

and V

→ V ,

then we can duplicate V .

Proof. By our assumptions,

E |(V

− V

) − (V

− V

→ 0

as n, m → ∞. So

− H

)dW

→ 0.

From our formulas for stochastic integrals, this means

− H

ds → 0.

This says that H

is a Cauchy sequence in the space L

(with respect to the norm k · k

given in Section 11). We will assume that you know that L

is complete or are willing to

believe it, so there exists H

such that

− H

ds → 0.

In particular H

→ H

, and this implies H

is adapted.

Let U

. Then as above,

E |(V

− V

) − U

= E

− H

)

ds → 0.

Therefore U

= V

− V

, and U has the desired form.

Next we show our result for a particular collection of options.

Proposition 16.2. If g is bounded,

g(W

) = c +

for an integrand H

that is adapted and some constant c.

Proof. By Ito’s formula with X

= −iuW

+ u

s/2 and f (x) = e

= 1 +

(−iu)dW

/2)ds

1
2

(−iu)

= 1 − iu

If we multiply both sides by e

−u

t/2

, which is a constant and hence adapted, we obtain

−iuW

= c

(16.2)

for an appropriate constant c

and integrand H

If f is a smooth function (e.g., C

∞

with compact support), then its Fourier trans-

form b

f will also be very nice. So if we multiply (16.2) by b

f (u) and integrate over u from

−∞ to ∞, we obtain

f (W

) = c +

for some constant c and some adapted integrand H. (We implicitly used Proposition 16.1,

because we approximate our integral by Riemann sums, and then take a limit.) Now using

Proposition 16.1 we take limits and obtain the proposition.

An almost identical proof shows that if f is bounded, then

f (W

− W

) = c +

for some c and H

Theorem 16.3. If V is F

measurable and E V

< ∞, then there exists a constant c and

an adapted integrand H

such that

V = c +

Proof. We will show that all V of the form

V = f

− W

) · · · f

− W

n−1

)

can be so represented. If we take linear combinations of these, then the linear combinations

can also be so represented. Since every V that is F

measurable can be written as a limit

of such linear combinations, Proposition 16.1 then implies the result.

The argument is by induction; let us do the case n = 2 for clarity. So we suppose

V = f (W

)g(W

− W

From Proposition 16.2 we now have that

f (W

) = c +

g(W

− W

) = d +

Set H

= H

if 0 ≤ s < t and 0 otherwise. Set K

= K

if s ≤ r < t and 0 otherwise. Let

= c +

and Y

= d +

. Then

hX, Y i

dr = 0.

Then by the Ito product formula,

= X

+ hX, Y i

= cd +

+ Y

]dW

If we now take s = u, that is exactly what we wanted. Note that X

+ Y

is 0 if

r > u; this is needed to do the general induction step.

17. Completeness.

Now let S

be a geometric Brownian motion. As we mentioned in the last section,

if S

= S

exp(σW

+ (µ − σ

/2)t), then given S

we can determine W

and vice versa, so

the σ fields generated by S

and W

are the same. Recall S

satisfies

= σS

+ µS

dt.

Define a new probability P by

dP
dP

= M

= exp(aW

− a

t/2).

By the Girsanov theorem,

= W

− at

is a Brownian motion under P. So

= σS

+ σS

adt + µS

dt.

If we choose a = −µ/σ, we then have

= σS

(17.1)

Since f

is a Brownian motion under P, then S

must be a martingale, since it is a

stochastic integral of a Brownian motion. We can rewrite (17.1) as

= σ

−1

(17.2)

Given a F

measurable variable V , we know by Theorem 16.3 that there exists

adapted H

such that

V = c +

But then using (17.2) we have

V = c +

−1

We have therefore proved

Theorem 17.1. If S

is a geometric Brownian motion and V is F

measurable, then there

exist a constant c and an adapted process K

such that

V = c +

Moreover, there is a probability P under which S

is a martingale.

The probability P is called the risk-neutral measure. Under P the stock price is a

martingale.

18. Black-Scholes formula, I.

We can now derive the formula for the price of any option. If V is F

measurable,

we have by Theorem 17.1 that

V = c +

(18.1)

and under P, the process S

is a martingale.

Theorem 18.1. The price of V must be E V .

Proof. This is the “no arbitrage” principle again. Suppose the price of the option V at

time 0 is W . Starting with 0 dollars, we can sell the option V for W dollars, and use the

W dollars to buy and trade shares of the stock. In fact, if we use c of those dollars, and

invest according to the strategy of holding K

shares at time s, then at time t we will have

− c) + V

dollars. At time t the buyer of our option exercises it and we use V dollars to meet that

obligation. That leaves us a profit of e

− c) if W

> c, without any risk. Therefore

must be less than or equal to c. If W

< c, we just reverse things: we buy the option

instead of sell it, and hold −K

shares of stock at time s. By the same argument, since

we can’t get a riskless profit, we must have W

≥ c, or W

= c.

Finally, under P the process S

is a martingale. So taking expectations in (18.1),

we obtain

E V = c.

Note that there is a slight difference with the approach we used in the binomial

asset model. There we showed that (1 + r)

−k

was a martingale under P. We could do

the analogue to that here, but it is slightly simpler to make S

a martingale under P and

to incorporate the interest rate into the definition of V . So, for example, in the case of

pricing a European call, we let

V = e

−rt

− K)

We can think of this as saying the value of V at time t is (S

− K)

in terms of the value

of the dollar at time t. In terms of present day dollars the value of V is e

−rt

− K)

The formula in the statement of Theorem 18.1. is amenable to calculation. Suppose

we have the standard European option, where V = e

−rt

− K)

. Recall that under P

the stock price satisfies

= σS

where f

is a Brownian motion under P. So then

= S

−σ

t/2

Hence

E V = E [e

−rt

− K)

]

(18.2)

= E

−rt

−(σ

/2)t

− K]

We know the density of f

is just (2πt)

−1

−y

, so we can do the calculations and end

up with the famous Black-Scholes formula:

= xΦ(g(x, t)) − Ke

−rt

Φ(h(x, t)),

where Φ(z) =

√

2π

−∞

−y

dy, x = S

g(x, t) =

log(x/K) + (r + σ

/2)t

√

h(x, t) = g(x, t) − σ

√

It is of considerable interest that the final formula depends on σ but is completely

independent of µ. The reason for that can be explained as follows. Under P the process S

satisfies dS

= σS

, where f

is a Brownian motion. Therefore, similarly to formulas

we have already done,

= S

−σ

and there is no µ present here. (We used the Girsanov formula to get rid of the µ.) The

price of the option V is

E e

−rt

− K]

(18.3)

which is independent of µ since S

is. Therefore we can use this instead of (18.2), i.e., we

can assume µ is zero, and the calculations become much simpler.

19. Black-Scholes formula, II.

Here is a second approach to the Black-Scholes formula. This approach works for

European calls and several other options, but does not work in the generality that the first

approach does. On the other hand, it allows one to computes what the equivalent strategy

of buying or selling stock should be to duplicate the outcome of the given option.

Let V

be the value of the portfolio and assume V

= f (S

, T − t) for all t, where f

is some function that is sufficiently smooth. We also want V

= (S

− K)

Recall Ito’s formula. The multivariate version is

f (X

) = f (X

) +

i=1

) dX

i,j=1

) dhX

, X

Here X

= (X

, . . . , X

) and f

denotes the partial derivative of f in the x

direction,

and similarly for the second partial derivatives.

We apply this with d = 2 and X

= (S

, T − t). From the SDE that S

solves,

= σ

dt, hX

= 0 (since T −t is of bounded variation and hence has no martingale

part), and hX

, X

= 0. Also, dX

= −dt. Then

− V

= f (S

, T − t) − f (S

, T − t)

(19.1)

, T − u) dS

−

, T − u) du

1
2

, T − u) du.

On the other hand,

− V

dβ

(19.2)

Since the value of the portfolio is

= a

+ b

we must have

= (V

− a

)/β

(19.3)

Also, recall

= β

(19.4)

We must therefore have

= f

, T − t)

(19.5)

and

r[f (S

, T − t) − S

, T − t)] = −f

, T − t) +

, T − t)

(19.6)

for all t and all S

. (19.6) leads to the parabolic PDE

1
2

+ rxf

− rf,

(x, s) ∈ (0, ∞) × [0, T ),

(19.7)

and

f (x, 0) = (x − K)

(19.8)

Solving this equation for f , f (x, T ) is what V

should be, i.e., the cost of setting up the

equivalent portfolio. Equation (19.5) shows what the trading strategy should be. In the

next section we show how to solve this PDE.

20. Solving PDE.

Without going into the theory of PDE, let us look at how to solve some simple PDE

using probability. Let us consider

= af

+ bf

f (x, 0) = g(x).

(20.1)

Here f is a function of x and t, a and b are given functions of x and g is also given. The

above equation is known as the Cauchy problem.

Proposition 20.1. Let A =

√

2a and let X

be the solution to

= A(X

)dW

+ b(X

)dt.

(20.2)

The solution to the above equation is given by

f (x, t) = E

g(X

Proof. Fix t

and let M

= f (X

, t

− t). We first show M

is a martingale. By Ito’s

formula,

= f

, t

− s)dX

− f

, t

− s)ds +

1
2

, t

− s)A

)ds

(20.3)

= f

, t

− s)A(X

)dW

+ f

, t

− s)b(X

)ds +

1
2

, t

− s)A

)ds

− f

, t

− s)ds.

Since f solves (20.1), then

= f

, t

− s)A(X

)dW

which is a stochastic integral of a Brownian motion, hence a martingale.

Now E

= f (x, t

) and E

= E

f (X

, 0) = E

g(X

). Since martingales

have constant expectation,

f (x, t

) = E

g(X

Since t

is arbitrary, the proposition is proved.

To solve the equation

= af

+ bf

+ cf,

f (x, 0) = g(x),

similar methods show that the solution is given by

f (x, t) = E

g(X

c(X

)ds

where X

is the solution to (20.2). (This is known as the Feynman-Kac formula.) To see

this, if we let

= M

c(X

)ds

where M

= f (X

, t

− t), then the Ito product formula yields

= M

c(X

)ds

c(X

)dt + e

c(X

)ds

Using (20.3) and the fact that af

+ bf

+ cf = 0, we see that that the dt term is 0 and

is a martingale. Using E N

= E N

leads to the desired representation of the solution.

Let us look at an example:

1
2

+ rxf

− rf,

which is the PDE that arises in Black-Scholes. Here a =

1
2

so that A = σx, b = rx,

and c = −r. The SDE to be solved, then, is

= σX

+ rX

dt,

= x.

We know the solution to this is

= xe

σW

−σ

t/2+rt

Hence

f (x, t) = E

g(X

−rt

= E

−rt

σW

−σ

t/2+rt

Since we know the density of W

, we can get an explicit expression (as an integral) for

f (x, t).

21. The fundamental theorem of finance.

In Section 17, we showed there was a probability measure under which S

was a

martingale. This is true very generally. Let S

be the price of a security. We will suppose

is a continuous semimartingale, and can be written S

= M

+ A

The NFLVR condition (“no free lunch with vanishing risk”) is that there do not

exist a fixed time T , ε, b > 0, and H

(that are adapted and satisfy the appropriate

integrability conditions) such that

(s) dS

> −

a.s.

for all t and

(s) dS

> b

> ε.

Here T, b, ε do not depend on n. The condition says that one can with positive

probability ε make a profit of b and with a loss no larger than 1/n.

Q is an equivalent martingale measure if Q is a probability measure, Q is equivalent

to P, and S

is a martingale under Q.

Theorem 21.1. If S

is a continuous semimartingale and the NFLVR conditions holds,

then there exists an equivalent martingale measure Q.

The proof is rather technical and involves some heavy-duty measure theory, so we

will only point examine a part of it. Suppose that we happened to have S

= W

+ f (t),

where f (t) is a deterministic function. To obtain the equivalent martingale measure, we

would want to let

= e

−

(s)dW

−

(s))

In order for M

to make sense, we need f to be differentiable. A result from measure

theory says that if f is not differentiable, then we can find a subset A of [0, ∞) such that
R

(s)ds = 0 but the amount of increase of f over the set A is positive. (This is not a

very precise statement.) Then if we hold H

= 1

(s) shares at time s, our net profit is

(s)dW

(s) df (s).

The second term would be positive since this is the amount of increase of f over the set

A. The first term is 0, since E (

(s)dW

)

(s)

ds = 0. So our net profit is

nonrandom and positive, or in other words, we have made a net gain without risk. This

contradicts “no arbitrage.”

Sometime Theorem 21.1 is called the first fundamental theorem of asset pricing.

The second fundamental theorem is the following.

Theorem 21.2. The equivalent martingale measure is unique if and only if the market is

complete.

We will not prove this.

22. American puts.

The proper valuation of American puts is one of the important unsolved problems

in mathematical finance. Recall that a European put pays out (K − S

)

at time T ,

while an American put allows one to exercise early. If one exercises an American put at

time t < T , one receives (K − S

)

. Then during the period [t, T ] one receives interest,

and the amount one has is (K − S

)

r(T −t)

. In today’s dollars that is the equivalent of

(K − S

)

−rt

. One wants to find a rule, known as the exercise policy, for when to exercise

the put, and then one wants to see what the value is for that policy. Since one cannot look

into the future, one is in fact looking for a stopping time τ that maximizes

E e

−rτ

(K − S

)

There is no good theoretical solution to finding the stopping time τ , although good

approximations exist. We will, however, discuss just a bit of the theory of optimal stopping,

which reworks the problem into another form.

Let G

denote the amount you will receive at time t. For American puts, we set

= e

−rt

(K − S

)

Our problem is to maximize E G

over all stopping times τ .

We first need

Proposition 22.1. If S and T are bounded stopping times with S ≤ T and M is a

martingale, then

E [M

| F

] = M

Proof. Let A ∈ F

. Define U by

U (ω) =

S(ω)

if ω ∈ A,

T (ω)

if ω /

∈ A.

It is easy to see that U is a stopping time, so by Doob’s optional stopping theorem,

E M

= E M

= E [M

; A] + E [M

; A

Also,

E M

= E M

= E [M

; A] + E [M

; A

Taking the difference, E [M

; A] = E [M

; A], which is what we needed to show.

Given two supermartingales X

and Y

, it is routine to check that X

∧ Y

is also a

supermartingale. Also, if X

are supermartingales with X

↓ X

, one can check that X

is again a supermartingale. With these facts, one can show that given a process such as

, there is a least supermartingale larger than G

So we define W

to be a supermartingale (with respect to P, of course) such that

≥ G

a.s for each t and if Y

is another supermartingale with Y

≥ G

for all t, then

≤ Y

for all t. We set τ = inf{t : W

= G

}. We will show that τ is the solution to the

problem of finding the optimal stopping time. Of course, computing W

and τ is another

problem entirely.

Let

= {τ : τ is a stopping time, t ≤ τ ≤ T }.

Let

= sup

τ ∈T

E [G

| F

Proposition 22.2. V

is a supermartingale and V

≥ G

for all t.

Proof. The fixed time t is a stopping time in T

, so V

≥ E [G

| F

] = G

, or V

≥ G

. so

we only need to show that V

is a supermartingale.

Suppose s < t. Let π be the stopping time in T

for which V

= E [G

| F

π ∈ T

⊂ T

. Then

E [V

| F

] = E [G

| F

] ≤ sup

τ ∈T

E [G

| F

] = V

Proposition 22.3. If Y

is a supermartingale with Y

≥ G

for all t, then Y

≥ V

Proof. If τ ∈ T

, then since Y

is a supermartingale, we have

E [Y

| F

] ≤ Y

= sup

τ ∈T

E [G

| F

] ≤ sup

τ ∈T

E [Y

| F

] ≤ Y

What we have shown is that W

is equal to V

. It remains to show that τ is optimal.

There may in fact be more than one optimal time, but in any case τ is one of them. Recall

we have F

is the σ-field generated by S

, and hence consists of only ∅ and Ω.

Proposiiton 22.4. τ is an optimal stopping time.

Proof. Since F

is trivial, V

= sup

τ ∈T

E [G

| F

] = sup

E [G

]. Let σ be a stopping

time where the supremum is attained. Then

≥ E [V

| F

] = E [V

] ≥ E [G

] = V

Therefore all the inequalities must be equalities. Since V

≥ G

, we must have V

= G

Since τ was the first time that W

equals G

and W

= V

, we see that τ ≤ σ. Then

E [G

] = E [V

] ≥ EV

= E G

Therefore the expected value of G

is as least as large as the expected value of G

, and

hence τ is also an optimal stopping time.

The above representation of the optimal stopping problem may seem rather bizarre.

However, this procedure gives good usable results for some optimal stopping problems. An

example is where G

is a function of just W

23. Term structure.

We now want to consider the case where the interest rate is nondeterministic, that

is, it has a random component. To do so, we take another look at option pricing.

Let r(t) be the (random) interest rate at time t. Let

β(t) = e

r(u)du

be the accumulation factor. One dollar at time T will be worth 1/β(T ) in today’s dollars.

Let V = (S

− K)

be the payoff on the standard European call option at time T

with strike price K, where S

is the stock price. In today’s dollars it is worth, as we have

seen, V /β(T ). Therefore the price of the option should be

β(T )

We can also get an expression for the value of the option at time t. The payoff, in terms

of dollars at time t, should be the payoff at time T discounted by the interest or inflation

rate, and so should be

−

r(u)du

− K)

Therefore the value at time t is

−

r(u)du

− K)

| F

= E

β(t)

β(T )

V | F

= β(t)E

β(T )

| F

From now on we assume we have already changed to the risk-neutral measure and

we write P instead of P.

A zero coupon bond with maturity date T pays $1 at time T and nothing before.

This is equivalent to an option with payoff value V = 1. So its price at time t, as above,

should be

B(t, T ) = β(t)E

β(T )

| F

= E

−

r(u)du

| F

Let’s derive the SDE satisfied by B(t, T ). Let N

= E [1/β(T ) | F

]. This is a

martingale. By the martingale representation theorem,

= E [1/β(T )] +

for some adapted integrand H

. So B(t, T ) = β(t)N

. Here T is fixed. By Ito’s product

formula,

dB(t, T ) = β(t)dN

+ N

dβ(t)

= β(t)H

+ N

r(t)β(t)dt

= β(t)H

+ B(t, T )r(t)dt,

and we thus have

dB(t, T ) = β(t)H

+ B(t, T )r(t)dt.

(23.1)

We now discuss forward rates. If one holds T fixed and graphs B(t, T ) as a function

of t, the graph will not clearly show the behavior of r. One sometimes specifies interest

rates by what are known as forward rates.

Suppose you want to borrow $1 at time T and repay it with interest at time T + ε.

At the present time we are at time t ≤ T . One can accomplish this by buying a zero

coupon bond with maturity date T and shorting (i.e., selling) B(t, T )/B(t, T + ε) zero

coupon bonds with maturity date T + ε. The value of the portfolio at time t is

B(t, T ) −

B(t, T )

B(t, T + ε)

B(t, T + ε) = 0.

At time T you receive $1. At time T + ε you pay B(t, T )/B(t, T + ε). The effective rate

of interest R over the time period T to T + ε is

εR

B(t, T )

B(t, T + ε)

Solving for R, we have

R =

log B(t, T ) − log B(t, T + ε)

We now let ε → 0. We define the forward rate by

f (t, T ) = −

∂

∂T

log B(t, T ).

(23.2)

Sometimes interest rates are specified by giving f (t, T ) instead of B(t, T ) or r(t).

Let us see how to recover B(t, T ) from f (t, T ). Integrating, we have

f (t, u)du = −

∂

∂u

log B(t, u)du = − log B(t, u) |

u=T
u=t

= − log B(t, T ) + log B(t, t).

Since B(t, t) is the value of a zero coupon bond at time t which expires at time t, it is

equal to 1, and its log is 0. Solving for B(t, T ), we have

B(t, T ) = e

−

f (t,u)du

(23.3)

Next, let us show how to recover r(t) from the forward rates. We have

B(t, T ) = E

−

r(u)du

| F

Differentiating,

∂

∂T

B(t, T ) = E

− r(T )e

−

r(u)du

| F

Evaluating this when T = t, we obtain

E [−r(t) | F

] = r(t).

(23.4)

On the other hand, from (23.3) we have

∂

∂T

B(t, T ) = −f (t, T )e

−

f (t,u)du

Setting T = t we obtain −f (t, t). Comparing with (23.4) yields

r(t) = f (t, t).

(23.5)

24. Some interest rate models.

Heath-Jarrow-Morton model

Instead of specifying r, the Heath-Jarrow-Morton model (HJM) specifies the forward

rates:

df (t, T ) = σ(t, T )dW

+ α(t, T )dt.

(24.1)

Let us derive the SDE that B(t, T ) satisfies. Let

∗

(t, T ) =

α(t, u)du,

∗

(t, T ) =

σ(t, u)du.

Since B(t, T ) = exp(−

f (t, u)du), we derive the SDE for B by using Ito’s formula with

the function e

and X

= −

f (t, u)du. We have

= f (t, t)dt −

df (t, u)du

= r(t)dt −

[α(t, u)dt + σ(t, u)dW

] du

= r(t)dt −

α(t, u)du

dt −

σ(t, u)du

= r(t)dt − α

∗

(t, T )dt − σ

∗

(t, T )dW

Therefore, using Ito’s formula,

dB(t, T ) = B(t, T )dX

1
2

B(t, T )(σ

∗

(t, T ))

= B(t, T )

r(t) − α

∗

1
2

(σ

∗

)

dt − σ

∗

B(t, T )dW

From (23.1)

dB(t, T ) = B(t, T )r(t)dt − σ

∗

B(t, T )dW

Comparing, we see that if P is the risk-neutral measure, we have α

∗

1
2

(σ

∗

)

If P is not the risk-neutral measure, it is still possible that one exists. Let θ(t) be a

function of t, let M

= exp(−

θ(u)dW

−

1
2

θ(u)

du) and define P(A) = E [M

; A] for

A ∈ F

. By the Girsanov theorem,

dB(t, T ) = B(t, T )

r(t) − α

∗

1
2

(σ

∗

)

+ σ

∗

θ]dt − σ

∗

B(t, T )df

where f

is a Brownian motion under P. So we must have

∗

1
2

(σ

∗

)

+ σ

∗

θ.

Differentiating with respect to T , we obtain

α(t, T ) = σ(t, T )σ

∗

(t, T ) + σ(t, T )θ(t).

If we try to solve this equation for θ, there is no reason off-hand that θ depends only on t

and not T . However, if θ does not depend on T , P will be the risk-neutral measure.

Hull and White model

In this model, the interest rate r is specified as the solution to the SDE

dr(t) = σ(t)dW

+ (a(t) − b(t)r(t))dt.

Here σ, a, b are deterministic functions. The stochastic integral term introduces random-

ness, while the a − br term causes a drift toward a(t)/b(t).

This is one of those SDE’s that can be solved explicitly. Let K(t) =

b(u)du.

Then

K(t)

r(t)

= e

K(t)

r(t)b(t) + e

K(t)

a(t) − b(t)r(t)

dt + e

K(t)

[σ(t)dW

]

= e

K(t)

a(t)dt + e

K(t)

[σ(t)dW

Integrating both sides,

K(t)

r(t) = r(0) +

K(u)

a(u)du +

K(u)

σ(u)dW

Multiplying both sides by e

−K(t)

, we have the explicit solution

r(t) = e

−K(t)

r(0) +

K(u)

a(u)du +

K(u)

σ(u)dW

If F (u) is deterministic, then

F (u)dW

= lim

F (u

)(W

i+1

− W

Linear combinations of Gaussian r.v.’s (Gaussian = normal) are Gaussian, and limits of

Gaussian r.v.’s are Gaussian, so we conclude

F (u)dW

is a Gaussian r.v. We see that

the mean at time t is

E r(t) = e

−K(t)

r(0) +

K(u)

a(u)du

We know how to calculate the second moment of a stochastic integral, so

Var r(t) = e

−2K(t)

2K(u)

σ(u)

du.

(One can similarly calculate the covariance of r(s) and r(t).) Limits of linear combinations

of Gaussians are Gaussian, so we can calculate the mean and variance of

r(t)dt and get

an explicit expression for

B(0, T ) = E e

−

r(u)du

Cox-Ingersoll-Ross model

One drawback of the Hull and White model is that since r(t) is Gaussian, it can take

negative values with positive probability, which doesn’t make sense. The Cox-Ingersoll-

Ross model avoids this by modeling r by the SDE

dr(t) = (a − br(t))dt + σ

r(t)dW

The difference from the Hull and White model is the square root of r in the stochastic

integral term. Provided a ≥

1
2

, it can be shown that r(t) will never hit 0 and will always

be positive. Although one cannot solve for r explicitly, one can calculate the distribution

of r. It turns out to be related to the square of what are known in probability theory

as Bessel processes. (The density of r(t), for example, will be given in terms of Bessel

functions.)

25. Foreign exchange.

Suppose we can buy bonds in dollars with constant interest rate r. So the value

of the bond at time t is B

= B

. Suppose we can buy bonds in pounds sterling with

constant rate u. If D

is the price of the bond, then D

= D

Let C

be the exchange rate; at time t one pound is worth C

dollars. Not surpris-

ingly, one model for exchange rates is to set

= C

σW

−σ

t/2+µt

where W

is a Brownian motion.

We want to see how to price options in terms of the exchange rate. There are two

main differences from the stock case. One is that if we buy pounds, we can invest them

in sterling bonds and receive interest. The other is that C

is not a quantity that one can

invest in. One cannot buy an exchange rate.

Let S

= C

. Start with with S

= C

dollars at time 0, exchange them

for S

pounds, and buy S

/(C

) sterling bonds. At time t redeem the bonds for

pounds each and convert back to dollars. So at time t, you have C

= S

dollars.

Therefore S

is tradable, even if C

is not.

The present value of S

is e

−rt

= B

−1

. Let us set Z

= B

−1

. Substituting

for C

and D

, we have

= e

−rt

σW

−σ

t/2+µt

= e

σW

−σ

t/2+(µ+u−r)t

Let M

= exp(−(µ + u − r)W

− (µ + u − r)

t/2) and define Q(A) = E [M

; A] if A ∈ F

By what are now standard calculations, Z

is a martingale under Q.

Now let’s price options. For example, one might have a sterling call, which is the

right at time T to buy a pound for K dollars. The payoff at time T is V = (C

− K)

We use the martingale representation theorem, just as in the Black-Scholes derivation, to

write

= c +

Since Z is a martingale under Q, then c = E

V and this is the value of the call in today’s

dollars. H gives the hedging strategy: at time s hold H

units of sterling bonds, and the

remainder, namely, V

− H

, in dollar bonds.

Are the calculations hard? Not really – the situation looks exactly the same as the

case of stock prices, but the interest rate is now r − u instead of r. This makes sense; we

don’t discount by the rate r, because insteading of holding stocks, which pay no interest,

we hold sterling bonds, which pay at interest rate u.

One could also do all these calculations from the point of view of the English in-

vestor. He has the alternative of putting his pounds into sterling bonds, or else exchanging

them for dollars and investing in dollar bonds. The calculations are very similar.

It turns out that the martingale measure (Q) will be different for the American and

the English investor. Nevertheless, they will come up with the identical value for an option

and will have the identical hedging strategy.

26. Dividends.

How do things change if a stock issues dividends? Of course, typically a stock issues

dividends at fixed discrete times, but for the sake of our continuous model, let us assume

that the stock issues dividends at a continuous rate.

There are two possibilities. One possibility is that the stock issues dividends at a

fixed rate, regardless of what the stock price is. This case is very similar to the foreign

exchange discussion: like the investor who exchanges his dollars for pounds and invests in

sterling bonds, the security purchased (sterling bonds in the foreign exchange case, stock

in the present case) pays interest.

So we will look at the other case, where we suppose that the stock issues dividends

at a rate proportional to the stock price. We suppose in a short time interval ∆t that the

amount of dividends issued is δS

∆t. Here δ is the dividend rate.

Suppose that as soon as we receive dividends, we immediately buy stock with it.

Our net gain in cash is then zero, but the number of shares we hold has increased. If S

the stock price, we hold e

δt

times as many shares, so our investment in the stock is now

= e

δt

= e

σW

−σ

t/2+µt+δt

We are back in the standard stock model, except that our mean rate of return has become

µ + δ instead of µ. The Black-Scholes formulas are much as before.

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